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TECHNICAL   MECHANICS. 


BY 


EDWARD  R.  MAURER, 

Professor  of  Mechanics  in  the  University  of  JVisconsitu 


SECOND  EDITION,  REVISED  AND  ENLARGED, 
SIXTH    THOUSAND 

OF    THE 

UNIVERSITY 

Of 

NEW  YORK : 

JOHN  WILEY  &  SONS. 

London:  CHAPMAN  &  HALL,  Limited. 
1910. 


4 


QBHEBAL 


Copyright,  1903, 


BY  'I 

SDWARD  R.  MAURBR.  1 


%«hrrt  9ntmtminii  anii  (Poaniaiig 
ISigw  lork 


PREFACE, 


This  work  is  not  a  pure  nor  an  applied  mechanics,  but  a  theo- 
retical mechanics  for  students  of  engineering,  and  in  accordance 
with  the  usage  of  some  German  writers,  the  title  "Technical 
Mechanics"  has  been  given  to  it. 

On  the  theoretical  side,  practically  all  the  subjects  treated 
have  a  direct  bearing  in  engineering  problems.  Little  atten- 
tion was  paid  to  experimentation,  principally  because  the 
author's  own  students  come  to  him  after  having  completed  a 
laboratory  course  in  elementary  mechanics.  In  general,  the 
theory  in  each  chapter  has  been  grouped  together  and  separated 
from  the  applications.  The  principal  equations  and  formulas 
are  set  in  bold-faced  type. 

On  the  applied  side,  no  attempt  was  made  to  present  fully 
any  one  subject,  as  the  analysis  of  trusses,  friction,  balancing 
of  rotating  systems,  etc.,  for  the  object  in  view  was  the  illus- 
tration of  the  use  of  principles  of  mechanics  and  not  treatments 
of  roof-trusses,  friction,  balancing,  etc.  However,  a  sufficient 
number  and  the  proper  kind  of  examples  have  been  included, 
it  is  believed,  to  give  to  the  student  a  working  knowledge  of 
the  subject. 

In  Statics,  especially,  a  distinctive  feature  is  the  nearly 
coextensive  use  of  graphical  and  algebraic  methods.  This  is 
due  to  the  author's  opinion  that  it  is  unwise  to  present  graphical 
and  analytical  statics  as  separate  courses,  at  least  to  beginners. 
The  treatments  of  composition  and  equilibrium  of  forces  are 
separate.  This  arrangement  is  largely  due  to  the  author's 
desire  to  develop  the  conditions  of  equilibrium  for  the  different 

Hi 


206178 


IV  PREFACE. 

classes  of  force  systems  consecutively  and  to  get  all  the  prin- 
ciples of  equilibrium  together.  The  examples  on  the  applica- 
tion of  the  principles  of  equilibrium  are  set  off  in  a  separate 
chapter,  number  VI.  No  attempt  was  made  to  arrange  them 
in  accordance  with  the  orders  of  Chapters  II  and  V.  On  the 
contrary,  such  an  arrangement  was  avoided,  and  the  entire  aim 
was  to  emphasize  the  fact  that  there  is  a  general  method  for 
solving  the  ordinary  problems  of  Statics  proper  and  that  it  con- 
sists principally  in  ** applying"  conditions  of  equilibrium. 

Kinematics  is  treated  mainly  as  a  preliminary  to  Kinetics, 
but  harmonic  motion  is  discussed  more  fully  than  usual  in 
works  of  this  kind. 

In  Kinetics,  D'Alembert's  Principle  is  used  considerably.  In 
spite  of  the  fact  that  it  has  been  described  as  clumsy  and  old- 
fashioned,  the  author  believes  that  the  use  of  the  principle  in 
all  but  simple  cases  is  decidedly  helpful,  and  besides  its  use 
makes  graphical  methods  possible. 

As  to  "mass  and  weight,"  what  has  been  called  the  physicist's 
usage  is  followed,  that  is,  mass  means  quantity  of  matter  and 
weight  the  Earth's  attraction.  The  author  retains,  however, 
the  familiar  equation  m  =  W^g,  and  as  holding  not  only  in 
gravitation  systems  of  units  but  in  all  systems  in  which  "unit 
force  gives  unit  mass  unit  acceleration."  Only  such  systems 
are  used  herein,  and  then  the  familiar  equation  F  =  ma  always 
holds.  This  of  course  requires  recognition  of  special  units  of 
mass  in  gravitation  systems.  The  author  has  succeeded  best 
with  his  classes  in  this  matter  by  doing  more  than  defining  these 
units — he  has  been  using  a  name  for  one  of  them,  the  "  English 
engineers'  unit"  (equal  to  32.2^  pounds).  So  strongly  does 
he  feel  that  names  are  helpful  in  this  instance  that  he  has  ven- 
tured to  put  them  into  print.  The  names  herein  used  are  "gee- 
pound"  and  "  geekilogram,"  denoting  3  2 .  2  ±  pounds  and  9 .  81  ± 
kilograms  respectively.  They  were  adopted  for  their  descrip- 
tiveness  and  as  better  than  "matt"  and  "ert.,"  the  only  other 
terms  proposed  in  this  connection  so  far  as  the  author  is  aware. 

For  the  information  of  any  instructor  who  may  consider 
using  this  book  as  a  text,  the  following  modifications  and  abridg- 
ments are  pointed  out:  Chapters  III  and  IV  may  follow  Chapter 


PREFACE.  V 

VI,  Chapter  V  may  be  taken  simultaneously  with  Chapter  II, 
and  Chapters  III,  IV,  and  XV,  and  parts  of  all  other  chapters 
may  be  omitted  without  serious  derangement  of  the  course. 

In  his  work  the  author  consulted  especially  the  books  of 
Profs.  L.  M.  Hoskins  and  John  Perry.  He  is  especially  indebted 
to  Prof.  C.  H.  Burnside  of  this  University  for  valuable  assistance 
on  Statics. 

Madison,  Wis.,  September,  1903. 

PREFACE  TO  THE   SECOND   EDITION. 


Several  additions  and  changes  have  been  made;  the  prin- 
cipal additions  are:  appendixes  E  and  F,  articles  87  and  D  15, 
and  an  explanation  of  principal  symbols,  page  409.  All  errors 
detected  in  the  first  edition  have  been  eliminated. 

September,  1904. 


TABLE  OF  CONTENTS. 


INTRODUCTION. 

ART. 

Nature  of  the  Subject i 

Division  of  the  Subject 2,  3 

Two  Methods  of  Analysis 4 

STATICS 

CHAPTER  I.    FORCE. 

1.  Preuminary. 

Force  Defined 5 

Action  and  Reaction 6 

Force  at  a  Distance  and  Force  by  Contact 7 

Distributed  and  Concentrated  Forces 8,  9 

Graphical  Representation  of  a  Force 10 

Notation 11 

2.  Measurement  of  Force;   Mass  and  Weight. 

Mass 12 

Units  of  Force 13 

Measurement  of  Force 14 

Weight 15 

3.  Force  Systems. 

Definitions 16 

Classification  of  Force  Systems 17 

CHAPTER  II.    EQUIVALENCE  OF  FORCE  SYSTEMS. 

I.  Preliminary. 

Definitions 18 

The  Principle  of  TransmissibiUty 19 

Graphical  Composition  of  Two  Concurrent  Forces 20 

Algebraic  Composition  of  Two  Concurrent  Forces 21 

Graphical    Composition    of    Three    Concurrent     Non-coplanar 

Forces 22 

Algebraic  Composition  of  Three  Concurrent  Non-coplanar  Forces 

Mutually  at  Right  Angles 23 

\  vii 


viii  TABLE  OF  CONTENTS. 


ART. 


Resolution  of  a  Force  into  Two  Concftrrent  Components 24 

Resolution  of  a  Force  into  Three  Non-coplanar  Forces 25 

Moment  of  a  Force  with  Respect  to  a  Point 26 

"  Varignon's  Theorem  " 27 

Moment  of  a  Force  with  Respect  to  a  Line 28 

Couples 29,  30 

Resolution  of  a  Force  into  a  Force  and  a  Couple 31 

2.  CoLUNEAR  Forces. 

Composition 32 

3.  CoPLANAR  Concurrent  Non-parallel  Forces. 

Graphical  Composition 33-35 

Algebraic  Composition 36 

4.  Coplanar  Non-concurrent  Parallel  Forces. 

Graphical  Composition 37-40 

The  Principle  of  Moments 41 

Algebraic  Composition 42,  43 

5.  Coplanar  Non-concurrent  Non-parallel  Forces. 

Graphical  Composition 44.  45 

The  Principle  of  Moments 46 

Algebraic  Composition 47 

Reduction  of  a  System  to  a  Force  and  a  Couple 48 

6.  Non-coplanar  Concurrent  Forces. 

Graphical  Composition 49 

Algebraic  Composition 50 

7.  Non-coplanar  Parallel  Forces. 

Graphical  Composition 51,  52 

The  Principle  of  Moments ~. 53 

Algebraic  Composition 54 

8.  Non-coplanar  Non-concurrent  Non-parallel  Forces. 

The  Resultant 55 

Graphical  Composition 56 

Principle  of  Moments 57 

Algebraic  Composition 58 

9.  Theory  of  Couples. 

Equivalent  Couples • 59 

Composition  of  Couples 6b 

Resolution  of  a  Couple 61 

CHAPTER  III.     CENTRE  OF  GRAVITY  AND  CENTROID. 

1.  Centroid  of  Parallel  Forces. 

Centroid  Defined 62 

Determination  of  the  Centroid 63 

2.  Centre  of  Gravity  of  a  Body. 

Definition  and  General  Formulas /. .  64 


TABLE  OF  CONTENTS.     '  ix 

ART. 

Moment  of  a  Weight  with  Respect  to  a  Plane 65 

Centre  of  Gravity  Determined  by  Integration 66 

Centre  of  Gravity  Determined  Experimentally 67 

3.  Centroids  of  Solids,  Surfaces,  and  Lines. 

Centroid  Defined 68,  69 

Moment  of  a  Volume,  Area,  or  Length 70 

Plane  of  Zero  Moment 71 

Centroids  of  Simple  Solids  and  Surfaces 72-80 

Centroids  of  Solids  and  Surfaces  Consisting  of  Simple  Parts 81 

Centroids  of  Solids  and  Surfaces  Considered  as  Parts  of  Other 

Solids  or  Surfaces 82 

Centroids  Determined  by  Integration 83-85 

Theorems  of  Pappus  and  Guldinus 86 

Centroid  Determined  Graphically ,     87 

CHAPTER  IV.     ATTRACTION    AND   STRESS. 

1.  Gravitation. 

Law  of  Gravitation 88 

Density 89 

Attraction  at  a  Point  or  "Strength  of  Field" 90 

Attractions  in  Some  Simple  Cases 91-97 

2.  Electric  and  Magnetic  Attractions. 

Laws  of  Electrostatic  and  Magnetic  Forces 98 

Strength  of  Field 99 

Analogy    between    Electrical    or    Magnetic    and    Gravitational 

Attractions 100 

Strengths  of  Field  Due  to  some  Simple    Distributions  of  Elec- 
tricity and  Magnetism loi 

3.  Stress. 

Stress  Defined 102 

Units  for  Expressing  Stress 103 

Classification  of  Stresses 104 

Description  of  a  Simple  Stress 105 

Intensity  of  Stress 106 

Graphical  Representation  of  a  Simple  Stress 107 

Centre  of  Stress. . 108 

A  Uniformly  Varying  Normal  Stress 109-1 1 2 

CHAPTER  V.     GENERAL  PRINCIPLES  OF  EQUILIBRIUM. 

I.  Preliminary. 

Definitions 113 

General  Condition  of  Equilibrium  of  a  System  of  Forces  Applied 

to  a  Rigid  Body 114 


X  TABLE  OF  CONTENTS. 

ART. 

Equilibrium  of  a  System  of  Forces  Apiflied  to  a  Non-rigid  Body.  .    115 
2.  The  Conditions  of  Equiubrium  for  thb  Various  Classes  of  Force 
Systems. 

CoUinear  Forces 116 

Coplanar  Concurrent  Non-parallel  Forces 117,   118 

Coplanar  Non-current  Parallel  Forces 119 

Coplanar  Non-concurrent  Non-parallel  Forces 120,   121 

'Non-coplanar  Concurrent  Forces 122 

Non-coplanar  Non-concurrent  Parallel  Forces 123 

Non-coplanar  Non-concurrent  Non-parallel  Forces 124 

Summary  of  Conditions  of  Equilibrium 125 

CHAPTER  VI.     APPLICATIONS   OF   THE   PRINCIPLES   OF 
EQUILIBRIUM. 

1.  Prbuminary. 

Nature  of  the  Problems 1 26 

General  Method  of  Solution 127 

2.  Flexible  Cords. 

Definitions 128 

Tension  in  a  Cord 1 29 

Position  Assumed  by  a  Cord  Sustaining  Loads 130,  131 

Position  Assumed  by  a  Heavy  Flexible  Cord  Suspended  from 

Two  Points 132,  133 

3.  Tackle. 

The  Pulley 134,  135 

4.  Smooth  Supports. 

Definitions 1 36 

Reaction  of  a  Smooth  Surface 137 

Pin  Joint  or  Hinge 138 

5.  Three  Typical  Problems  on  Coplanar  Non-concurrent  Forces. 

Problem  1 139 

Problem  II 140 

Problem  III 141 

6.  Jointed  Frames. 

Definitions 142 

The  Pin  Pressures 143 

General  Direction  for  Solving  Examples 144 

7.  Jointed  Frames — Continued. 

Kind  of  Frames  Considered 145 

" Force  or  Stress  in  a  Member" * 146 

Method  for  Determining  the  Force  or  Stress  in  a  Member 147 

Graphical  Method  for  "Analyzing"  Trusses 148-151 

8    Rough  Supports;  Friction. 

Definitions 152-155 


TABLE  OF  CONTENTS.  xi 

ART. 

Laws  of  Friction 156 

Determination  of  Coefficient  of  Friction 157 

Friction  Circle 158 

Cone  of  Friction 159 

Belt  Friction 160 

"Forces  in  Space"  and  Miscellaneous. 

Examples  Involving  Non-parallel  Non-coplanar  Forces 161 

Miscellaneous 162 


KINEMATICS. 

CHAPTER  VII.    RECTILINEAR  MOTION  OF  A  PARTICLE. 

ART. 

I.  Velocity  and  Acceleration. 

Specification  of  Position 163 

Space-time  Curve 1 64 

Displacement 165 

Kinds  of  Rectilinear  Motion 166 

Velocity 167-169 

Velocity-time  Cur\'e 170 

Velocity  Increment 171 

Kinds  of  Non-uniform  Motion 172 

Acceleration 173-175 

"  Acceleration-time  "  and  other  Curves 176 

II.  Important  Special  Motions. 

Uniform  Motion 177 

Uniformly  Accelerated  Motion 178 

Simple  Harmonic  Motion 179,  180 

Motion  of  the  Piston  of  a  Steam-engine i8i 

CHAPTER  VIII.     CURVILINEAR  MOTION. 

I.  Velocity  and  Acceleration. 

Specification  of  Position 182 

Space-time  Curve 183 

Displacement 184 

Kinds  of  Motion 185 

Velocity 186 

Speed-time  Curve  and  Hodograph 187 

Acceleration 188 

II.  Resolutions  of  Velocities  and  Acceler.a.tions. 

Components  and  Resultant  of  Velocities  or  Accelerations  Defined 189 

Axial  Components  of  a  \''elocity igo 

Tangential  and  Normal  Components  of  Velocity 191 


xn  Ty4BLE  OF  CONTENTS. 

ART. 

Axial  Components  of  Acceleration t 192 

Tangential  and  Nonnal  Components  of  Acceleration 193 

in.  Relativity  of  Motion. 

Meaning  of  Relative  Path,  Displacement,  Velocity,  and  Acceleration...  194 

Definitions 195 

Relation  between  the  Velocities  and  Accelerations  of  a  Point 196 

Meaning  of  Composition  of  Motions 197 

IV.  Composition  of  Simple  Harmonic  Motions. 

Mechanism  for  Compounding  Simple  Harmonic  Motions 198 

Composition  of  Two  Collinear  S.  H.  M.'s  of  Equal  Periods 199 

Resolution  of  a  S.  H.  M.  into  Two  Components 200 

Composition  of  Many  Collinear  S.  H.  M.'s  of  Equal  Periods 201 

Composition  of  Two  S.  H.  M.  's  in  Lines  at  Right  Angles 202 

Resolution  of  a  S.  M.  H.  into  Two  Rectangular  Components 203 

Composition  of  more  than  Two  S.  H.  M.'s  not  Collinear 204 

CHAPTER  IX.     MOTION  OF  A  RIGID  BODY. 

I.  Translation. 

Translation  Defined 205 

Motions  of  all  Points  of  a  Body  in  Translation  are  alike 206 

Velocity  and  Acceleration  of  the  Body 207 

II.  Rotation. 

Rotation  Defined 208 

Angular  Displacement 209 

Angular  Velocity 210-212 

Angular  Acceleration 213-215 

Velocity  and  Acceleration  of  any  Point  of  a  Rotating  Body 216 

III.  Any  Plane  Motion. 

Plane  Motion  Defined 217 

Angular  Displacement 21& 

Angular  Velocity  and  Angular  Acceleration 219 

Velocity  and  Acceleration  of  any  Point  of  the  Body 220 

Plane  Motion  Regarded  as  a  Combined  Translation  and  Rotation 221 

Instantaneous  Axis  (of  no  Velocity) 222 

Instantaneous  Rotation '. 223 


KINETICS. 

CHAPTER  X.    MOTION   OF  A   PARTICLE   (RESUMED)   AND    OF  A 
SYSTEM  OF  PARTICLES. 

I.  Mass  and  Mass-centre. 

Quantity  of  Matter ■. 224 

Mass >. ...  225 

Practical  Determination  of  Mass ; - 226 


TABLE  OF  CONTENTS.  xiii 

ART. 

Moment  of  Mass 227 

Mass-centre  Defined 228 

Relation  between  Mass-centre  and  Centre  of  Gravity 229 

II.  Motion  of  a  Particle. 

Laws  of  Motion 230 

Quantitative  Expression  of  the  Second  Law  of  Motion 231 

Kinetic  System  of  Units 232 

Relations  between  Force  Units  and  between  Mass  Units 233 

Relation  between  Mass  and  Weight  of  a  Body 234 

Acceleration  of  a  Particle  Acted  upon  by  Several  Forces 235 

Equations  of  Motion  of  a  Particle 236 

III.  Motion  of  a  System  of  Particles. 

Definitions 237 

D'Alembert's  Principle 23S 

Component  of  an  Effective  System  Along  any  IJne 239 

Motion  of  the  Mass-centre  of  any  System  of  Particles 24c 

Moment  of  the  Effective  System  About  any  Axis 241 

"Angular  Motion"  of  a  System  of  Particles 242 

CHAPTER  XI.    TRANSLATION  OF  A  RIGID  BODY  (RESUMED). 

I.  General  Principles. 

Equations  of  Motion 243 

Resultant  of  the  Effective  System 244 

II.  Applications. 

General  Method  of  Procedure 245 

Kinetic  Reactions 246 

Vibrations 247-25 1 

Kinetic  Friction 252 

CHAPTER  XII.     ROTATION  (RESUMED). 

I.  Second  Moments  of  Mass  (Moment  of  Inertia,  etc.) 

Occurrence  of  Second  Moments 253 

Moment  of  Inertia 2  54 

Radius  of  Gyration 255 

Relations  between  Moments  of  Inertia  and  between  Radii  of  Gyration 

with  Respect  to  Parallel  Axes 256 

Composite  Bodies , 257 

Experimental  Determination  of  Moment  of  Inertia 258 

Product  of  Inertia 1 259 

Principal  Axes 260 

II.  General  Principles. 

The  Effective  Forces 261 

Moment  of  the  Effective  System 262 

Equations  of  Motion 263 

Resultant  of  the  Effective  System • 264,  265 


xiv  TABLE  OF  CONTENTS. 

ART. 

III.  Applications. 

Determination  of   the  Motion •. , 266 

Motion  of  Pendulums 267 

Torsion  Balance 268 

Conical  Pendulum 269 

Weighted  Conical  Pendulum  Governor 270 

Kinetic  Reactions 271-274 

Balancing  of  Rotating  Bodies 275-277 

Pivot  and  Journal  Friction 278 

CHAPTER  XIII.    ANY    PLANE    MOTION    OF    A    RIGID    BODY 

(RESUMED). 

I.  General  Principles. 

The  Effective  Forces 279 

Moment  of  the  Effective  System ^ 280 

Equations  of  Motion 281 

Resultant  of  the  Effective  System 282 

II.  Applications. 

Determination  of  the  Motion 283 

Kinetic  Reactions 284 

Rolling  Resistance 285 

CHAPTER  XIV.     WORK  AND  ENERGY. 

I.  Work. 

Work  Defined 286 

Expressions  for  Work  Done  by  a  Force 287 

Sign  of  Work 288 

Unit  of  Work * '. 289 

Work  Diagram —  - 290 

Work  Done  by  Gravity  upon  a  Body  in  any  Motion 291 

Work  Done  by  Concurrent  Forces  and  by  their  Resultant 292 

Work  Done  by  a  Pair  of  Equal,  Opposite,  and  CoUinear  Forces 293 

Work  Done  by  a  Body  Against  a  Force 294 

II.  Energy. 

Energ}'  Defined 295 

Kinetic  Energy  Defined 296 

Kinetic  Energy  of  a  Particle.  ." 297 

Kinetic  Energy  of  any  System  of  Particles 298 

Potential  Energy  Defined 299 

The  Amount  of  Potential  Energy 300 

Potential  Energy  of  a  System  not  Always  a  Definite  Quantity....  301-303 

Localization  of  Potential  Energy 304 

Other  Forms  of  Energy 305 

III.  Principles  of  Work  and  Energy. 

Principles  of  Work  and  Kinetic  Energy 306 

Principle  of  Work  and  Energy  for  Conser-'ative  Systems 307 


Tj4BLE  of  contents,  XV 

ART. 

Conservation  of  Energy 308 

Principle  of  Energy  for  Machines 309-311 

IV.  Applications. 

Computation  of  Velocity  and  Distance 312 

Train  Resistance 313 

Friction  Brakes 314 

Efficiency  of  Tackle 315 

Efficiency  of  a  Mine-hoist 316 

CHAPTER  XV.    IMPULSE  AND  MOMENTUM. 

I.  Impulse. 

Impulse  of  a  Force  whose  Direction  is  Constant 317 

Impulse  of  a  Force  whose  Direction  Varies 318 

Component  of  an  Impulse 319 

Moment  of  the  Impulse  of  a  Force 320 

II.  Momentum. 

Momentum  of  a  Particle 321 

Components  of  a  Momentum 322 

Moment  of  Momentum 323 

Momentum  of  a  System  of  Particles 324 

Moment  of  the  Momentum  of  a  System  of  Particles 325 

Momentum  of  a  Rigid  Body  in  Special  Cases 326 

III.  Principles  of  Impulse  and  Momentum. 

Principles  for  a  Particle 327 

Principles  for  a  System  of  Particles 328 

rv.  Applications. 

Computation  of  Velocity  and  Time 329 

Pressures  Due  to  Jets 330 

Sudden  Impulses 331 

Force  of  a  Blow  and  Recoil  of  a  Gun 332 

Collision  or  Impact 333~335 

Ballistic  Pendulum  and  Centre  of  Percussion 336 


APPENDIX   A. 

\TCTORS. 

Scalar  and  Vector  Quantities Ai 

Vector  Defined A2 

Addition  of  Vectors A3 

Negative  of  a  Vector A4 

Subtraction  of  Vectors A5 


*vi  TABLE  OF  CONTENTS. 

APPENDIX  ^B. 

RATES. 

ART 

Kinds  of  Variable  Quantities B  i 

Rate  of  a  Uniform  Scalar ■ B  2 

Rate  of  a  Non-uniform  Scalar B3 

Sign  of  a  Rate B  4 

Unit  of  Rates B5 

Rate  of  a  Uniform  Vector E  6 

Rate  of  a  Non-uniform  Vector B  7 

Descriptive  Terms B8 


APPENDIX   C. 

DIMENSIONS  OF  UNITS. 

Magnitude  of  a  Quantity Cl 

Fundamental  and  Derived  Units C  2 

Dimensions  of  Units C3 

Application  of  the  Theory  of  Dimensions ,  C4 


APPENDIX   D. 

SECOND  MOMENTS  OF  AREA. 

I.  Moment  of  Inertia. 

Moment  of  Inertia  Defined , D  i 

Units  of  Moment  of  Inertia D  2 

Radius  of  Gyration D  3 

Relations  between  Moments  of  Inertia  and  between  Radii  of  Gyration.     D  4 
Composite  Areas D  5 

n.  Product  of  Inertia. 

Product  of  Inertia  Defined D  6 

Units  of  Product  of  Inertia.  .  .■ D  7 

Zero  Products  of  Inertia. DS 

Relations  between  Products  of  Inertia D9 

Composite  Areas D  10 

in.  Relations  between  Moments  of  Inertia  with  Respect  to  Axes 
Inclined  to  Each  Other. 

General  Equations D  11 

Geometrical  Constructions D  12,  13 

Principal  Axes  and  Moments  of  Inertia • D14 

Graphical  Determination  of  Moment  oi  Inertia D  15 


TABLE  OF  CONTENTS.  xvii 

APPENDIX    E. 
VIRTUAL  WORK. 

ART. 

Definitions E  i 

Principle  of  Virtual  Work  for  a  Particle .-  E  2 

Principle  of  Virtual  Work  for  a  System  of  Particles E  3 

Application  of  the  Principles  of  Virtual  Work  to  Statically  Indeter- 
minate Problems E  4 


APPENDIX   F. 

SUPPLEMENT  TO  STATICS. 

Truss  Loads F  1-5 

Computation  of  Apex  Loads F  6 

Determinations  of  Reactions F  7-9 

Maximum  Stresses F  10 

Classification  of  Frames F  11 

Moments  of  Forces  Determined  Graphically F  12-14 

To  "  Pass"  a  Funicular  Polygon  through  Three  Points P  15 

Relation  between  Two  Funicular  Polygons  for  a  Given  Force  System 

Drawn  from  the  Same  Pole F  16 

To  Close  a  Gauche  Polygon  with  Three  Sides  whose  Directions  are 

Given ' F  17 


OF    THe 

UNIVERSITY 

OF 


TECHNICAL  MECHANICS. 


INTRODUCTION. 

1.  Nature  of  the  Subject. — Mechanics,  broadly  defined,  is 
the  science  which  treats  of  motion.  It  is  a  natural  science,  and 
not  a  branch  of  mathematics  as  the  student  is  apt  to  infer  from 
his  observation  that  mathematics  is  extensively  used  in  the 
subject.  Its  foundations  consist  in  a  few  grand  generalizations 
from  experience,  such  as  Newton's  Laws  of  Motion;  the  super- 
structure consists  in  the  deduction  of  the  logical  consequences 
of  those  generalizations. 

2.  Division  of  the  Subject. — The  following  divisions  may 
be  made: 

,  V  ,,    V      .       ( Kinematics 
r        (I)  Mechanics    |  (Kinetics 

■  (Dynamics    |  g^^^j^^ 

Kinematics  treats  of  methods  for  the  description  of  motion. 
Dynamics  deals  with  the  circumstances  of  motions,  such  as  their 
causes,  etc.  Kinetics  embraces  that  part  of  dynamics  which 
deals  with  variable  motion,  and  Statics  that  part  dealing  with 
uniform  motion.* 

The  above  classification  and   order  is   usually  followed  in 
modem  works  on  pure  mechanics,  and  is  no  doubt  the  logical 
one. 
^  r  Rigid  Solids 

(.)  Mechanics  of  ]^fS°"'^« 
f  1^  Gases 

The  general  principles  are  the  same  for  these  latter  branches, 
but  there  are  special  ones  and  special  methods  for  each  branch ; 

*  In  a  treatment  of  mechanics  on  the  plan  outhned,  it  would  develop 
that  the  principles  relating  to  bodies  at  rest  are  included  in  Statics. 


2  INTRODUCTION. 

thus  it  is  convenient  to  treat  them  separately.  The  present 
volume  deals  with  principles  commen  to  all  these  branches,  but 
relates  especially  to  the  first  one.  It  is  the  practice  in  most 
American  schools  to  present  the  mechanics  of  non-rigid  solids, 
liquids,  and  gases  in  separate  courses,  the  first  two  under  the 
titles  of  Strength  of  Materials  and  Hydraulics  respectively,  and 
the  last  in  connection  with  Thermodynamics. 

3.  Technical  Mechanics. — By  this  term  is  meant  a  presenta- 
tion of  the  general  principles  of  mechanics  with  special  refer- 
ence to  their  application  in  the  fields  of  engineering. 

Although  the  order  of  treatment  outlined  in  the  first  division 
of  the  subject  (art.  2)  is  the  logical  one,  another  is  followed 
herein.  This  is  the  historical  order:  Statics,  Kinematics,  and 
Kinetics.  Statics  as  presented  also  occupies  a  more  important 
part  than  in  the  outline  above,  and  is  not  limited  by  the  pre- 
ceding definition;  it  deals  especially  with  principles  and  appli- 
cations relating  to  bodies  at  rest. 

Since  the  logical  order  is  not  followed,  several  principles 
appear  in  the  early  pages  which  are  not  fully  explained  or  justi- 
fied until  later. 

4.  Two  Methods  of  Analysis. — They  are  called  the  graphical 
and  the  algebraical,  or  analytical. 

In  the  first  method,  the  quantities  under  consideration  are 
represented  by  lines  and  the  analysis  is  wholly  by  means  of 
geometrical  figures.  In  calculations  by  this  method  leading  to 
numerical  results  the  figures  are  accurately  drawn  to  scale. 
In  the  second  method  the  quantities  under  consideration  are 
represented  by  symbols,  and  the  analysis  is  by  ordinary  algebra. 
Calculations  are  carried  out  by  arithmetic. 

The  graphical  method  finds  its  best  application  in  statics, 
though  it  is  advantageous  in  the  applications  of  kinematics  to 
mechanisms.  Statics  is  often  treated  entirely  by  one  of  the 
two  methods,  such  a  treatment  being  known  as  graphical  or 
analytical  statics,  as  the  case  may  be;  in  this  book  both  methods 
are  employed.  Some  discussions  and  solutions  are  given  by 
both  methods  by  way  of  comparison,  the  others  by  that  method 
which  is  the  more  suitable. 


STATICS. 

CHAPTER  I. 
FORCE. 

§  I.     Preliminary. 

5.  Force  Defined. — Bodies  act  upon  each  other  in  various 
ways,  producing  different  kinds  of  results.  Those  actions 
which  influence  the  motion  of  bodies  are  now  to  be  considered. 

Definition. — An  action  of  one  body  upon  another  which 
changes  or  which  exerted  alone  would  change  the  state  of  the 
motion  of  the  body  acted  upon  is  called  a  force. 

We  say  that  a  force  is  applied  to  a  body,  a  force  is  exerted 
upon  a  body,  a  force  acts  upon  a  body,  etc.  The  last  expression 
is  faulty  and  misleading;  for,  since  force  is  the  name  for  a  cer- 
tain kind  of  an  action,  the  expression  might  be  rendered  thus: 
an  action  acts  upon  a  body.  Now  it  is  not  an  action  which  acts, 
but  some  other  body.  The  expression  criticised  is,  however, 
a  current  one  and  is  used  in  this  book. 

Our  first  notions  about  force  are  founded  on  our  experience 
with  forces  exerted  by  or  upon  ourselves.  From  this  experi- 
ence we  have  learned  that  a  force  has  magnitude,  direction,  and 
place  of  application. 

6.  Action  and  Reaction. — When  one  body  exerts  a  force 
upon  another,  the  latter  also  exerts  a  force  upon  the  former, 
and  the  two  forces  are  equal  in  magnitude  but  opposite  in 
direction.  This  fact  is  often  referred  to  as  the  principle  of 
''action  and  reaction''  the  phrase  being  an  abbreviation  of 
Newton's  third  law  of  motion.  By  "action"  is  meant  one  of 
the  forces,  and  by  "reaction"  the  other. 

7.  Force  at  a  Distance  and  Force  by  Contact. — This  is  a 
classification  for  convenience  and  is  probably  not  based  on  fact. 

Gravitational,   electrical,   and   magnetic   forces   have   been 

3 


4  FORCE.  [Chap.  I. 

called  actions  at  a  distance,  the  force  between  the  bodies  con- 
cerned being  exerted  without  their  Contact  and,  as  was  formerly- 
supposed,  without  any  material  connection  between  them.  It 
is  now  known  that  the  force  between  two  electrified  bodies 
depends  upon  the  medium  in  which  they  are  placed,  which  is 
proof  that  the  medium  has  to  do  with  the  exertion  of  the  force 
by  one  body  upon  the  other.  Even  gravitation,  it  is  be- 
lieved, is  not  a  true  action  at  a  distance,  and  "the  earth  and 
a  stone  do  not  strictly  draw  each  other  together,  but  are 
pushed  together  by  something  which  extends  from  one  to  the 
other."  But  apparently  these  forces  are  actions  at  a  distance, 
and  they  will  be  so  called. 

Pressure  of  the  atmosphere  upon  any  object,  the  force  of  a 
hammer  blow  upon  a  nail,  etc.,  are  forces  by  contact.  The 
place  of  application  of  a  contact  force  is  the  surface  of  contact 
between  the  bodies  concerned. 

8.  Distributed  and  Concentrated  Forces. — This  is  a  classifi- 
cation for  convenience  and  is  not  in  accord  with  fact. 

A  distributed  force  is  one  whose  place  of  application  is  a  surface 
or  a  solid.  For  example,  the  water  pressure  on  a  ship,  the  place 
of  application  being  the  wetted  surface;  the  attraction  of  the 
earth  upon  a  stone,  the  place  of  application  being  the  solid 
defined  by  the  stone;  etc.  All  forces  are  really  distributed 
forces. 

A  concentrated  force  is  one  of  finite  magnitude  whose  place 
of  application  is  a  point.  Such  a  force  is  imaginary  since  no 
actual  (finite)  force  can  be  applied  at  a  point,  but  there  are 
actual  forces  whose  place  of  application  is  very  small  and  which 
may  be  regarded  as  concentrated  forces.  The  conception  is 
useful  in  developing  principles  relating  to  actual  forces.  Thus, 
as  just  stated,  many  actual  forces  are  practically  concentrated 
and  may  be  treated  as  such,  and  a  distributed  force  is  regarded 
as  consisting  of  a  great  number  of  concentrated  forces. 

The  line  of  action,  or  action  line,  of  a  concentrated  force  is  a 
line  parallel  to  its  direction  and  containing  its  point  of  applica- 
tion. Note  the  distinctions  between  the  terms  action  line, 
direction,  and  sense.  To  illustrate,  imagine  a  pull  exerted  upon 
a  sled  by  means  of  a  single  cord.     The  action  line  of  the  force  is 


§L]  PRELIMINARY.  5 

the  line  determined  by  the  taut  cord,  its  direction  is  upward 
and  to  the  right  30°  with  the  horizontal,  and  its  sense  is  upward 
and  to  the  right,  not  downward  and  to  the  left. 

9.  Specification  of  a  Concentrated  Force. — A  concentrated 
force  is  completely  specified  by  its  magnitude,  direction,  and 
application  point.  It  is  explained  later  that  the  effect  of 
such  a  force  applied  to  a  rigid  body  does  not  depend  upon  its 
application  point,  but  only  on  its  magnitude,  action  line,  and 
sense,  which  are  therefore  called  the  essential  characteristics  of 
a  force  (as  regards  rigid  bodies). 

10.  Graphical  Representation  of  a  Force.  —  Since  a  force 
has  magnitude  and  direction,  it  is  a  vector  quantity;*  the 
magnitude  and  direction  of  a  force  may  therefore  be  represented 
by  a  vector — the  length  of  the  vector  repre- 
senting to  some  scale  the  magnitude  of  the  - 
force,  and  the  direction  of  the  vector  giving 
the    direction    of   the    force.     If  the  force  be 

A  B 

concentrated,  the    vector  may  also  represent  > 

the  action  line,  but  it  is  convenient  to  represent  ^^^'  ^• 

the  action  line  separately.  Thus  suppose  that  the  irregular 
outline  in  fig.  1  represents  a  body  to  which  a  force  is  applied 
at  P  horizontally  to  the  right.  The  vector  AB  represents 
the  magnitude  and  direction  of  the  force,  and  a  horizontal  line, 
indefinite  in  length  through  P,  its  action  line. 

If  the  vector  were  drawn  through  P  it  would  serve  also  to 
locate  the  action  line,  but  representation  by  two  lines  is  con- 
venient especially  when  many  forces  applied  to  the  same  body 
are  under  consideration.  The  part  of  the  entire  figure  in  which 
the  body  is  represented  and  the  action  lines  are  drawn  is  called 
the  space  diagram,  and  the  part  in  which  the  vectors  are  drawn 
is  called  the  vector  diagram. 

11.  Notation. — In  the  graphical  analysis,  each  vector  and 
the  corresponding  action  line  will  be  marked  by  the  same  letters 
a  capital  at  each  end  of  the  vector  and  the  same  small  letters, 
one  on  each  side,  at  the  action  line  as  in  fig.  i.  Reference  to  a 
force  in  the  text  will  be  by  capital  letters;  thus  ''the  force  AB'' 

*  See  Appendix  A  for  brief  discussion  of  vectors. 


6  FORCE,  [Chap.  L 

means  one  whose  magnitude  and  directicn  are  represented  by 
AB  and  whose  action  line  is  ah.       * 

In  the  algebraic  analysis,  a  force  will  be  denoted  by  a  capital 
letter,  which  will  also  stand  for  the  magnitude  of  the  force. 

§  II.    Measurement  of  Force;  Mass  and  Weight. 

12.  Mass. — By  mass  of  a  body  is  meant  the  quantity  of 
matter  in  it.  There  are  several  independent  units  of  mass  in 
use;  only  two  will  be  described  here. 

The  "British  unit"  is  the  quantity  of  matter  in  a  certain 
piece  of  platinum  deposited  in  the  Office  of  the  Exchequer, 
London ;  this  unit  is  called  a  pound. 

The  "French  unit"  is  the  quantity  of  matter  in  a  certain 
piece  of  platinum  deposited  in  the  Palais  des  Archives,  Paris; 
this  unit  is  called  a  kilogram. 

Copies,  more  or  less  exact,  of  these  standards  and  multiples 
and  submultiples  of  them  are  in  common  use  especially  for  the 
measurement  of  quantity  of  matter  in  trade. 

13.  Units  of  Force. — A  gravitation  unit  of  force  is  a  force 
equal  to  the  Earth's  attraction  on  a  unit  mass. 

Gravitation  units  are  not  constant  with  regard  to  place,  for 
the  Earth's  attractions  on  equal  masses  at  different  places  are, 
in  general,  not  equal;  in  fact,  the  attractions  are  in  the  ratio  of 
the  values  of  g  in  the  formula 

^  =  32.0894  (i +0.0052375  sin2/)(i  — 0.0000000957  ^) 

computed  for  the  two  places,  /  denoting  latitude  and  e  eleva- 
tion above  sea  level  in  feet.  The  extreme  variation  is  between 
the  attractions  at  a  high  elevation  on  the  equator  and  at  the 
pole;  this  variation  is  but  0.6  per  cent,  while  for  points  within 
the  United  States  the  maximum  variation  is  about  0.3  per  cent. 
Ordinarily  no  account  need  be  taken  of  the  differences  in  the 
gravitation  units  of  force  employed  at  different  places,  for  errors 
introduced  into  engineering  calculations  by  this  variation  are 
practically  always  negligible. 

It  is  customary  to  call  any  gravitation  unit  of  force  and  the 


§11.]  MEASUREMENT  OF  FORCE.  "7 

unit  mass  on  which  it  is  based  by  the  same  name.*     Correspond- 
ing to  the  above-described  units  of  mass  we  have  then 

the  unit  of  force  called  a  pound,  which  is  a  force  equal  to 

the  Earth's  attraction  on  a  pound  mass,  and 

the  unit  of  force  called  a  kilogram,^h.\Qh  is  a  force  equal 

to  the  Earth's  attraction  on  a  kilogram  mass. 

An  absolute  unit  of  force  is  one  whose  value  is  independent 
of  time  or  place.     Several  such  units  are  described  later. 

14.  Measurement  of  Force. — The  lever  balance  is  primarily 
a  force-measuring  device,  and  it  is  the  one  commonly  used  for 
measuring  forces.  The  force  to  be  measured  is  applied  at  one 
end  of  a  lever  or  system  of  levers,  and  the  Earth's  attraction  on 
a  body  of  known  mass,  m  pounds  say,  at  the  other,  the  mass 
being  such  that  the  two  forces  balance.  If  the  ratio  between 
two  such  forces  (determined  once  for  all  by  the  balance-maker) 
is  w,  then  the  magnitude  of  the  force  being  measured  is  nm 
pounds.  As  the  reader  knows,  the  balance-maker  provides  that 
the  numerical  value,  nm,  may  be  read  off  directly.  It  should  be 
noticed  that  a  lever  balance  measures  forces  in  terms  of  a  gravi- 
tation unit. 

By  means  of  lever  balances,  engineers  measure  forces  applied 
to  materials  in  testing  their  strength,  and  usually  also,  with  a 
slight  indirection,  the  "brake  resistance"  when  testing  engines 
and  other  motors. 

The  spring  balance  is  another  force-measuring  device.  The 
force  to  be  measured  is  applied  to  the  spring  so  as  to  stretch  it 
merely.  The  force  corresponding  to  each  amount  of  stretch, 
within  the  range  of  the  spring,  having  been  determined  by  the 
maker,  the  magnitude  of  the  force  being  measured  may  be  in- 
ferred from  the  stretch.  As  the  reader  well  knows,  the  maker 
provides  that  the  numerical  value  of  the  force  may  be  read  off 
directly.     In  theory  at  least,   forces  measured  with   a  spring 

*  Such  usage  is  apt  to  confuse  the  beginner,  and,  when  necessary  for 
clearness,  we  shall  add  an  explanatory  word;  thus,  'pound  mass'  or 
'pound  force.'  The  word  '  second'  is  used  in  a  closely  analogous  way. 
It  is  the  name  for  two  different  units,  one  of  angle  and  the  other  of  time, 
and  the  latter  is  defined  with  reference  to  the  first.  To  be  clear,  we 
often  say  second  of  arc  or  second  of  time,  as  the  case  may  be. 


8  FORCE.  [Chap.  L 

balance  are  measured  in  terms  of  a  gravitation  unit  for  the  place 
at  which  it  was  graduatea.  For,  consider  the  theory  of  the 
graduation  of  a  spring  balance:  a  body  of  known  mass,  a  pound 
say,  is  suspended  from  it,  and  the  position  of  the  pointer  on  the 
spring  is  scribed  on  the  scale ;  the  position  is  marked  one  pound 
(force),  for  it  corresponds  to  a  stretch  due  to  a  pound  force; 
then  other  bodies  whose  masses  are  multiples  or  submultiples 
of  the  unit  mass  are  successively  hung  from  the  balance  and  the 
scale  is  marked  correspondingly. 

For  measuring  some  forces  the  spring  balance  is  better 
adapted  than  the  other.  Engineers  use  it  to  measure  the  pres- 
sure of  the  steam  in  a  running  engine,  the  pull  of  a  locomotive, 
etc. 

15.  Weight. — By  weight  of  a  body  will  be  meant  the  Earth's 
attraction  upon  it.  Weight  as  here  defined  is  a  force  and  must 
be  measured  in  force  units.  It  is  customary  to  express  weights 
in  gravitation  units. 

As  before  stated,  the  weight  of  any  body  changes  slightly 
with  change  of  its  locality.  However,  if  it  is  expressed  in  a 
gravitation  unit  the  numerical  value  of  the  weight  remains  the 
same,  for  the  relative  changes  in  the  weight  and  the  unit  are 
equal.  An  analogy  is  the  measurement  of  the  length  of  an  iron 
rod  by  a  standard  of  the  same  material  at  two  different  times, 
the  temperature  having  changed  during  the  interval.  The 
length  of  the  rod  has  changed,  but  the  numerical  value  as  deter- 
mined by  the  iron  standard  remains  constant,  for  the  relative 
changes  in  the  lengths  of  the  rod  and  standard  are  the  same. 
If  the  weight  of  a  body  is  expressed  in  an  absolute  unit,  the 
numerical  value  will  change  just  as  the  weight  changes  if  the 
body  be  transported. 

A  lever  balance,  then,  will  give  the  same  numerical  value  for 
the  weight  of  a  body  at  all  places,  while  a  spring  balance,  if 
sufficiently  sensitive,  will  show  the  true  variation  in  its  weight. 


§111.]  FORCE  SYSTEMS.  9 

§  III.     Force  Systems. 

1 6.  Definitions. — Any  number  of  forces  collectively  con- 
sidered is  called  a  system  of  forces  or  a  force  system. 

The  forces  of  a  system  are  called  coplanar  when  their  action 
lines  are  in  the  same  plane,  and  non-coplanar  when  they  are  not 
in  a  plane. 

The  forces  of  a  system  are  called  concurrent  when  their  action 
lines  intersect  at  a  point,  and  non-concurrent  when  they  do  not 
so  intersect. 

The  forces  of  a  system  are  called  parallel  when  their  action 
lines  are  parallel,  and  non-parallel  when  they  are  not  parallel. 

Force  systems  may  be  described  in  accordance  with  the  above 
definitions,  thus:  concurrent  systems,  non-coplanar  parallel 
systems,  etc.,  according  as  the  forces  of  the  system  are  con- 
current, non-coplanar  and  parallel,  etc. 

A  system  of  two  forces  which  are  equal,  parallel,  opposite  in 
sense,  and  have  different  action  lines  is  called  a  couple, 

17.  Classification  of  Force  Systems. — A  classification  may 
be  made  in  various  ways.  In  Chaps.  II  and  V  the  treatment  is 
based  upon  the  following  classification: 

,n  .  i  Parallel 

Concurrent  j  Non-parallel 

Coplanar  {  ^  r»      iT  t 

(  Non-con-current  j  J,^^^"^^    ,,  , 
^  ( Non-parallel 

(  Concurrent 
Non-coplanar "{  ^^  f  Parpllpl 

^  (  Non-concurrent  \  f  ^raiiei 

^  (  Non-parallel 


CHAPTER  II. 
EQUIVALENCE  OF  FORCE  SYSTEMS. 

§  I.     Preliminary. 

1 8.  Definitions. — Equivalent  force  systems  are  such  as  may 
be  substituted  for  each  other  without  change  of  effect. 

The  resultant  of  a  force  system  is  the  simplest  equivalent 
system.  The  resultant  of  a  system  acting  upon  a  rigid  body 
consists  of  either  one  or  two  forces,  as  will  be  shown  later.  It 
follows  from  the  above  definitions  that  two  equivalent  systems 
have  the  same  resultant.  The  components  of  a  force  are  any 
forces  whose  resultant  is  that  force. 

Composition  of  a  force  system  is  the  process  of  finding  a 
simpler  equivalent  system.  Finding  the  resultant  of  a  sy^em 
is  the  most  important  case  of  composition.  Resolution  of  a 
force  system  is  the  process  of  finding  a  less  simple  equivalent 
system.  Finding  components  of  a  force  is  the  most  important 
case  of  resolution. 

19.  Principle  of  Transmissibility.  —  The  effect  of  a  force 
applied  to  a  rigid  body  is  the  same  for  all  application  points  in 
its  action  line.  This  follows  from  the  equations  of  motion  of  a 
rigid  body  (art,  242)^  for  they  are  independent  of  the  application 
points  of  the  applied  forces. 

This  principle  may  be  roughly  verified  by  experiment  with 
the  apparatus  represented  in  fig.  2,  which  consists   of   a  body 
I  suspended   from    two    spring    balances. 
The  springs  are  elongated  on  account  of 
the  weight  of  the  body,  and  if  a  force, 
]   as  F,  be   applied  at  A ,  the  springs  will 


If  suffer  additional  elongations  which  in  a 

Fig.  2.  way  are  a  measure  of  the  effect  of  the 

applied  force.     If  the  application  point  of  F  be  changed  to  B 


§1.] 


PRELIMINARY. 


II 


or  C,  the  spring  readings  will  not  change,  hence  the  effect  of  F 
has  not  changed. 

20.   Graphical   Composition    of   Two  Concurrent   Forces. — 

The  Parallelogram  Law. — If  two  forces  acting  upon  a  rigid 
body  be  represented  by  OA  and  OB,  then  their  resultant  is 
represented  by  the  diagonal  OC  of  the  parallelogram  OABC 
(see  fig.  3). 


Fig. 

1k 
3. 

The  Triangle  Law. — If  two  concurrent  forces  acting  upon  a 
rigid  body  be  represented  in  magnitude  and  direction  by  AJ5 
and  BC,  their  resultant  is  represented  in  magnitude  and  direc- 
tion by  the  side  AC  of  the  triangle  ABC  (see  fig.  4). 

The  second  law  is  obviously  a  consequence  of  the  first,  and 
the  parallelogram  law  m^y  be  verified  experimentally  as  follows : 
Set  a  drawing  board  in  a  vertical  position  and  fasten  a  spring 
balance  and  smoothly  working  pulleys  to  it  somewhat  as  shown 
in  fig.  5.  Next  fasten  three  cords  to  a  small  ring,  tie  their 
loose  ends  to  the  hook 
of  the  balance  and  to 
two  bodies  whose 
weights  are  known 
and  lay  two  cords  w.fl 
over  the  pulleys  as 
shown. 

The  strings  now 
exert  three  forces  on  the  ring  (F^,  F^,  and  F3)  equal  to  the 
weights  (1^1  and  W^  of  the  suspended  bodies  and  the  reading 
(5)  of  the  balance,  respectively.  Since  F^  "balances"  F^  and 
F2,  the  resultant  of  F^  and  F^  must  be  equal  and  directly  opposite 
to  F3.  We  have  only  to  ascertain  now  whether  a  construction 
according  to  the  parallelogram  law  gives  a  resultant  equal  and 


Fig.  5. 


12 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


opposite  to  Fy  So  lay  off  on  the  board  OA  and  OB  equal  (by- 
some  scale)  to  F^  and  F^,  and  ^complete  the  parallelogram 
OABC.  Then  OC  should  (according  to  the  law)  represent  the 
resultant  of  F^  and  F^,  i.e.,  it  should  represent  a  force  equal  and 
opposite  to  Fg  and  it  will  be  found  that  it  does. 


EXAMPLES. 

1.  The  large  square  (fig.  6)  represents  a  board  3X3  feet 
upon  which  five  forces  are  applied  as  shown.  Determine  com- 
pletely the  resultant  of  the  4-  and  5-lb.  forces. 

2.  Determine  completely  the  resultant  of  the  7-  and  8-lb. 
forces. 

Q 


8  lbs. 


'^  4  lbs. 


6  lbs 


(a) 


Fig.  6. 

21.  Algebraic    Composition    of  Two  Concurrent  Forces. — 

Let  P  and  Q  (fig.  7)  be  two  concurrent  forces  and  a  the  angle 
between  their  action  lines.  Of  course  there  are  two  angles 
here;  the  one  taken  is  that  between  the  parts  of  the  lines 
on  the  sides  of  their  intersection  toward  which  the  arrows 
point. 

The  lines  AB  and  5C  represent  the  magnitude  and  direction 
of  P  and  Q  respectively;  then  AC  represents  the  magnitude  and 
direction  of  their  resultant.  The  action  line,  parallel  to  ACy  is 
marked  i^      Since  the  angle  C5Z>  =  a, 


and 


AC^  =^AB^  +  BC^ -^2AB  BC  cos  a. 


tan  CAD  =  BC  sin  a/{AB+BC  cos  a). 


§!•] 


PRELIMINARY, 


13 


If  R  denotes  the  resultant  and  6  the  angle  between  R  and  P, 
then,  since  AC  represents  the  value  of  R  and  the  angle  CAD 
equals  6, 

R2  =  p2+Q2+2PQ  cosa (l) 

and 

tan  ^  =  Q  sin  a:/(P  +  Q  cosa).       ...     (2) 

Special   cases:     If    a  =  90°,   i^=(P2+g2^i  and  tan  d=Q/P, 
Describe  the  resultant  if  a=o°;  if  a  =  180°. 


B 

a' 

'  y 

0 

.>-^^^ 

D 

.-' 

/ 

( 

^ 

{ 

3' 

c 


Fig.  8. 


EXAMPLES. 

1.  Solve  ex.  i,  art.  20,  by  the  formulas  of  this  article. 

2.  Compute  the  resultant  of  the  6-  and  7 -lb.  forces,  fig.  6. 

3.  Show  how  the  magnitude  of  the  resultant  of  two  forces 
P  and  Q  changes  as  a  changes  from  0°  to  180°. 

22.  Graphical  Composition  of  Three  Concurrent  Non- 
coplanar  Forces. — Parallelopiped  of  Forces. — If  three  non-co- 
planar  forces  acting  upon  a  rigid  body- 
be  represented  by  OA,  OB,  and  OC,  their 
resultant  is  represented  by  the  diagonal 
OD  of  the  parallelopiped  O^^C-D.  (See 
fig.  8.) 

Proof  :    According  to  the  parallelo- 
gram law  OC^  represents  the  resultant 
of  two  of  the  forces,  OD  represents  the 
resultant  of  the  third  force  and  OC  and  hence  of  the  three 
given  forces. 

From  an  inspection  of  the  figure,  it  is  plain  that  the  magni- 
tude and  direction  of  the  resultant  of  three  non-coplanar  con- 
current forces  is  given  by  their  vector  sum. 

23.  Algebraic  Composition  of  Three  Concurrent  Non-Coplanar 
Forces  Whose  Action  Lines  are  Mutually  at  Right  Angles. — 
Denote  the  three  forces  and  their  resultant  by  P,  Q,  5,  and  R, 
and  the  acute  angles  between  P  and  R,  Q  and  R,  and  5  and 
R,  by  ^1,  6^,  and  6^,  respectively.  Supposing  that  P,  Q,  and  S 
are  represented  by  OA,  OB,  and  OC  of  fig.  8,  it  is  plain  that 

R2  =  P2  +  Q2+S2 

and 

cos  6^  =  P/R,    cos  62  =  Q/R,    cos  6^  =  S/R. 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


EXAMPLES. 

1.  Suppose  that  a  force  of  9  Ibs/acts  at  A  (fig.  6),  perpen- 
dicularly to  the  plane  of  the  board  and  outward.  Determine 
the  resultant  of  the  9-,  4-,  and  7-lb.  forces. 

2.  Change  the  sense  of  the  9-lb.  force  and  solve  ex.  i. 

24.  Resolution  of  a  Force  into  Two  Concurrent  Components. 
— It  may  be  performed  by  applying  the  triangle  or  parallelo- 
gram law  inversely.  Thus,  let  it  be  required  to  resolve  the 
force  AB  (fig.  9),  applied  at  P.  Draw  from  A  and  B  two  lines 
which  intersect  in  any  point  C;  then  AC  and  CB  represent 
the  magnitudes  and  directions  of  components  of  AB.  The 
action  lines  of  the  components  must  intersect  on  ab,  and  their 
appUcation  points  must  be  rigidly  connected  with  P.  For,  if 
the  two  forces  AC  and  CB  be  compounded,  their  resultant  will 
be  found  to  be  AB. 


(a)  (b) 

Fig.  9.  Fig.  10. 

The  problem  just  solved  is  indeterminate,  for,  C  being  any 
point,  there  are  an  infinite  number  of  solutions.  If  conditions 
are  imposed  upon  the  components,  the  resolution  is  more  or  less 
definite;  for  example,  if  in  the  above  it  had  been  specified  that 
the  components  should  be  horizontal  and  vertical,  there  would 
be  but  one  answer. 

Rectangular  Components,  or  Resolved  Parts. — An  important 
case  of  resolution  is  that  in  which  the  angle  between  the  com- 
ponents is  90°.  Each  is  called  a  rectangular  component,  or 
resolved  part,  of  the  force.  They  can  always  be  readily  com- 
puted.    From  fig.  10,  it  is  plain  that 

the  rectangular  com-  "1  fthe  cosine  of  the 
ponent  or  resolved  !  _  j  the  magnitude  of  )  J  acute  angle  be- 
part  of  a  force  along  f  ~"  (  that  force  )  1  tween  the  force 
any  line                          J  [      and  that  line. 

If  the  rectangular  components  of  a  force  F  are  parallel  to 
coordinate  axes,  as  x  and  y,  they  are  called  x-  and  ^/-components 


§1-3 


PRELIMINARY. 


15 


of  F  respectively,  and  will  be  denoted  by  Fx  and  Fy.  If  the 
acute  angles  between  the  force  and  the  x  and  y  axes  be  denoted 
by  a  and  /?  respectively, 


and 


Fa;  =  FC0SQ: 

Fy  =  F  cos /?  =  F  sin  a. 


EXAMPLES. 

1.  Resolve  the  5-lb.  force  of  fig.  6,  page  12,  into  two  com- 
ponents whose  action  lines  are  parallel  to  the  4-  and  6-lb. 
forces  respectively. 

2.  Resolve  the  5-lb.  force  of  fig.  6  into    two    components 

^  whose  action  lines  are  parallel  to  the  8-  and  6-lb.  forces. 
3.  Resolve  the  8-lb.  force  of  fig.  6  into  two  components  one 
of  which  is  horizontal  and  the  other  6.5  lbs.  in  magnitude. 

4.  Resolve  the  6-lb.  force  of  fig.  6  into  two  components 
whose  lines  of  action  are  horizontal  and  vertical  respectively. 

5.  Resolve  the  4-  and  7-lb.  forces  of  fig.  6  into  horizontal  and 

t  vertical  components. 
25.  Resolution  of  a  Force  into  Three  Non-coplanar  Forces, — It 
'  may  be  performed  by   applying  the   parallelopiped   of  forces 
inversely.     Thus,  let  it  be  required  to  resolve  the  force  repre- 
sented  by    OD  (fig.  11).      Construct   a 
parallelopiped   of   which   OD  is  a  diag- 
onal.    The  three  edges  intersecting  at 
O  represent  the  components  of  OD;  the 
application  points  must  be   rigidly  con- 

^nected  to  that  of  the  given  force. 
The  problem  just  solved  is  indeter- 
minate, for  any  number  of  such  paral- 
lelopipeds  may  be  thus  drawn.  If 
conditions  are  imposed  upon  the  com- 
ponents, the  resolution  is  more  or  less 

r  definite. 

I  Rectangular  Components. — An  important  case  of  resolution 
is  that  in  which  the  three  components  are  mutually  at  right 
angles.     Each  is  a  rectangular  component,  for  the  two  com- 


y 

B 

/^ 

I             / 

\ 

/ 


X 


Fig. 


1 6  EQUiyALENCE  OF  FORCE  SYSTEMS.  [Chap.  IL 

ponents  of  OD  (fig.  ii),  one  of  which  is  either  OA,  OB,  or  OC, 
are  at  right  angles  to  each  other,  as*OA  and  OA'. 

If  the  three  components  of  a  force  F  are  parallel  respectively 
to  coordinate  axes  x,  y,  and  z,  they  are  called  %-,  y-,  and  0-com- 
ponents  of  F,  and  will  be  denoted  by  F^,  Fy,  and  F^  respec- 
tively. If  the  acute  angles  between  the  force  F  and  the  x,  y, 
and  z  axes  be  denoted  by  a,  /?,  and  f  respectively, 

F;,;  =  F  cos  a,     Fy =F  cos  /?,     F^  =  F  cos  7-. 

In  some  instances,  it  may  be  more  convenient  to  determine  these 
components  as  follows:  First  resolve  the  given  force  into  two 
rectangular  components  one  of  which  is  parallel  to  one  of  the 
axes;  the  other  will  be  parallel  to  the  plane  of  .the  other  two 
axes.  Then  resolve  the  second  component  into  two  forces 
w^hich  are  parallel  to  these  two  axes.  For  example,  if  OD 
(fig.  11)  be  resolved  first  along  the  x  axis,  the  first  resolution 
gives  OA  and  OA';  the  resolution  of  OA^  gives  OB  and  OC. 
Also, 

OA  =0D  cos  a,  or     Fx  =  F  cos  a; 

OB  =  04'  cos  /?'  =  OD  sin  a  cos  /9',     or     Fy  =  F  sin  a  cos  /?' ; 

OC  =  Oyl' cos  ;''  =  0i)  sin  a  cos  ;-',     or     F^  =  i^  sin  a  cos  7''. 

EXAMPLE. 

Resolve  each  force  acting  on  the  cube  in  fig.  3S(c)  into  its  x, 
y,  and  z  components. 

26.  Moment  of  a  Force  with  Respect  to  a  Point,  —  The 
moment  of  a  force  with  respect  to  a  point  is  the  product  of  the 
magnitude  of  the  force  and  the  perpendicular  distance  between 
its  action  line  and  the  point.  The  perpendicular  distance  is 
called  the  arm  of  the  force  with  respect  to  that  point,  and  the 
point  is  called  an  origin  or  a  centre  of  moments. 

In  the  following,  the  moment  of  a  force  will  usually  be  de- 
noted by  M,  and  the  moment  of  a  force  with  respect  to  an  origin 
0  by  M^. 

The  moment  of  a  force  with  respect  to  a  point  is  a  measure 
of  its  tendency  to  rotate  the  body  upon  which  it  acts  about  that 
point.  For,  if  the  body  is  fixed  at  that  point  but  free  to  turn 
about  it  in  a  given  plane,  any  force  in  that  plane  will  cause  it 


§1.] 


PRELIMINARY. 


17 


to  rotate  about  the  fixed  point.  The  amount  of  this  tendency 
is  proportional  to  the  magnitude  of  the  force  and  to  its  arm  with 
respect  to  that  point,  and  hence  to  the  moment  of  the  force  with 
respect  to  the  point. 

The  Unit  Moment. — From  the  definition  of  moment  of  a 
force,  it  follows  that  the  unit  moment  is  the  moment  of  a  unit 
force  whose  arm  is  a  unit  length ;  hence  there  are  many  units  of 
moments.  We  have  no  short  names  for  any  of  them,  but  they 
are  called  a  foot-pound,  inch- ton,  etc.,  according  as  the  units  of 
length  and  force  are  the  foot  and  pound,  or  inch  and  ton,  etc. 

The  Sign  of  a  Moment. — Sign,  plus  or  minus,  is  given  the 
moment  of  force  according  as  it  produces  or  tends  to  produce 
counter-clockwise  or  clockwise  rotation  about  the  origin  of 
moments.  In  the  following,  the  rotation  is  supposed  to  be 
viewed  from  the  reader's  side  of  the  printed  page. 

27.  "Varignon*s    Theorem."  —  The    algebraic   sum    of   the 
moments  of  two  concurrent  forces  with  respect  to  an  origin  in 
their   plane   equals   the   moment  of  their       p 
resultant  with  respect  to  that  origin. 

Proof :  Suppose  that  the  vectors  marked 
P  and  Q  and  R  (fig.  12)  represent  two 
concurrent  forces  and  their  resultant 
respectively.  Call  the  arms  of  the  three 
forces  with  respect  to  O,  p,  q,  and  r  and 
the  angles  between  the  action  lines  and  a 
perpendicular  to  OA,  a,  /?,  and  0  respectively, 
figure,  it  is  plain  that 

P  cos  a+Q  cos  /?  =  i^  cos  6; 


therefore       P  0.4  cos  a  -h  Q  OA  cos  /? =i?  OA  cos  6, 

Pp  +  Qq=^Rr. 


or 


Q.E.D. 


Supply  proof  when  0  is  between  P  and  R,  or  Q  and  R. 

According  to  the  theorem,  the  moment  of  a  force  equals  the 
sum  of  the  moments  of  its  x  and  y  components ;  often  it  is  easier 
to  compute  this  sum  than  the  moment  directly.  When  one 
component  passes  through  the  origin,  the  moment  of  the  other 
equals  that  of  the  force. 


i8 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  U 


EXAMPLE. 

Compute  the  moment  of  the  6-lb.  force  (fig.  6)  about  C 
directly  and  from  its  horizontal  and  vertical  components. 

28.  Moment  of  a  Force  with  Respect  to  a  Line. — If  a  foice 
be  resolved  into  components  parallel  and  perpendicular  to  a 
given  line,  the  product  of  the  magnitude  of  the  perpendicular 
component  and  the  distance  from  its  action  line  to  the  -given 
line  is  called  the  moment  of  the  force  with  respect  to  the  line. 
The  line  is  called  an  axis  of  moments^  and  the  distance  referred 
to  above  is  called  the  arm  of  the  perpendicular  component  with 
respect  to  that  axis.  Thus,  suppose  F  (Fig.  13)  is  a  force 
acting  upon  a  body,  not  shown,  and  AC  and  AB  are  its  com- 
ponents parallel  and  perpendicular  respectively  to  the  moment 
axis  OY.  Then  if  YL  is  the  perpendicular  between  01"  and 
AB,  the  moment  of  F  with  respect  to  OY  is  AB-  YL,  or  F^YL. 


.    Fig.  13. 

The  value  of  the  moment  does  not  depend  on  the  point  A 
selected  at  which  the  force  is  resolved,  for  obviously  the  value 
of  Fj  does  not  depend  on  A,  and  neither  does  the  value  of  YL, 
since  YL  equals  the  distance  between  OY  and  a  plane  parallel 
to  it  through  F,  which  distance  is  clearly  independent  of  A . 

Some  special  <:ases:  (i)  When  the  force  is  parallel  to  the 
axis,  its  moment  is  zero;  (2)  when  the  force  intersects  the 
axis,  its  moment  is  zero;    (3)  when  the  force  is  perpendicular 


§  I.]  PRELIMINARY.  1 9 

to  the  axis,  its  moment  is  the  product  of  the  force  and  the 
distance  between  it  and  the  axis. 

The  moment  of  a  force  with  respect  to  a  line  is  a  measure  of 
its  tendency  to  rotate  the  body  on  which  it  acts  about  that  line. 
For,  if  a  body  is  free  to  turn  about  the  line,  OF  say,  any  force, 
as  AD,  acting  upon  it  will  cause  it  to  rotate.  Now  the  com- 
ponents of  AD  would  produce  the  same  effect  upon  the  body  as 
does  AD,  but  a  component  parallel  to  the  axis  would  produce 
no  rotation  and  therefore  the  rotation  effects  oi  AD  and  the 
perpendicular  component,  AB,  are  the  same.  But  the  ten- 
dency oi  AB  to  produce  rotation  is  proportional  to  AB  and  its 
arm,  hence  to  their  product,  or  the  moment  of  ^D. 

The  sign  of  the  moment  of  a  force  with  respect  to  a  line  is 
taken  as  positive  or  negative  according  as  the  force  produces 
counter-clockwise  or  clockwise  rotation  about  the  axis  of  mo- 
ments. The  sign  will  depend  upon  the  side  from  which  the 
rotation  is  viewed.  If  the  axis  is  also  a  coordinate  axis,  the 
rotation  is'  customarily  viewed  from  its  positive  end. 

A  second  method  to  compute  the  moment  of  a  force  with 
respect  to  an  axis:  Resolve  the  force  into  three  rectangular 
components  one  of  which  is  parallel  to  the  axis ;  then  compute 
the  moments  of  the  other  two  components  with  respect  to 
that  axis,  and  add  these  moments  algebraically;  this  sum  is 
the  moment  of  the  force  itself.  Thus,  by  this  method,  the  moment 
of  F  about  OF  is  F^-Ax  —  Fz-Az.  That  this  method  and 
the  first  give  equivalent  results  can  be  shown  thus :  the  moment 
of  the  force  by  the  first  method  F^  •  YL  is  also  the  moment  of 
F^  about  F,  and  the  moment  by  the  second  method  is  the 
algebraic  sum  of  the  moments  of  the  components  Fx  and  F^  of 
Fj  about  Y'y  hence,  by  Varignon's  theorem, 

F,'YL=F^'A^-F,-Az. 

EXAMPLE. 

Compute  the  moments  of  each  force  of  fig.  38  {c)  with 
respect  to  the  x,  y,  and  z  axes,  die  edges  of  the  cube  being  4  ft. 
long. 

29.  Couples. — Definitions. — Two  equal  and  opposite  forces 
not  coUinear  are  called  a  couple.     By  arm  of  a  couple  is  meant 


\ 


20  EQUiyALENCE  OF  FORCE  SYSTEMS.  [Chap   IL 

the  distance  between  the  action  lines  of  its  forces.  The  moment 
of  a  couple  with  respect  to  a  point'in  its  plane  is  the  algebraic 
sum  of  the  moments  of  its  forces  with  respect  to  that  point. 

By  the  definition,  the  sign  of  the  moment  of  a  couple  must 
be  the  same  as  the  sign  of  the  algebraic  sum.  of  the  moments  of 
its  forces.  The  sign  of  the  sum  is  the  same  for  all  origins  (see 
proposition  below)  and  can  be  seen  at  a  glance  for  an  origin 
between  the  forces  or  on  the  action  line  of  one  of  them.  The 
sense  of  a  couple  refers  to  the  sign  of  its  moment.  We  speak  of 
positive  and  negative  sense  or  counter-clockwise 
and  clockwise  senses.  By  aspect  of  a  couple  is 
meant  the  aspect  *  of  its  plane. 
^*^  e'i^^'^  *^  Proposition. — The  moment  of  a  couple  is  the 
same  for  all  origins,  and  it  equals  the  product  of 
^o'"  the  magnitude  of  one  of  the  forces  of  the  couple 
Fig.  14.  and  the  arm.  Proof:  The  moments  of  the  couple 
of  fig.  14  with  respect  to  O',  O",  and  O'"  are  respectively, 


F-0"A+F-0''B  =  F'AB, 


and  F'0"'A-'F'0'"B=F'AB. 

Since  0',  O",  and  0'"  represent  all  possible  origins  in  the  plane, 
the  proposition  is  proved. 

30.  Graphic  Representation  of  a  Couple. — The  moment  of  a 
couple,  its  aspect,  and  its  sense  may  be  represented  by  a  vector. 
The  length  of  the  vector  is  made  equal,  by  some  scale,  to 
the  moment  of  the  couple,  it  is  drawn  normal  to  the  plane  of 
the  couple,  and  its  arrow  is  made  to  correspond  with  the  sense 
of  the  couple.  This  correspondence  depends  upon  some  arbi- 
trary rule,  such  as  the  following:  the  arrow  on  the  vector  points 
toward  the  place  from  which  the  rotation  of  the  couple  appears 
counter-clockwise.  The  vector  thus  representing  a  couple  will 
for  brevity  be  called  the  vector  of  the  couple. 

For  example,  the  couple  in  the  upper  face  of  the  parallelo- 

*  Aspect  of  a  plane  refers  not  to  its  position,  but  to  its  direction;   it 
is  conveniently  specified  by  the  direction  of  a  line  normal  to  it. 


51.] 


PRELIMINARY. 


21 


Vector  scale:   1  in. =100  ft.  lbs. 
Fig.   15. 


piped  of  fig.  15  is  represented  by  the  vector  AB  or  CD.     Each 

vector    also    represents     any    other 

couple    whose    moment,   aspect,   and 

sense   are  the  same  as  the  moment, 

aspect,  and  sense  of  the  upper  couple; 

the  couple  in  the  lower  face  is  such 

a  one. 

It  is  shown  in  art.  59  that  couples 
whose  moments,  aspects,  and  senses 
are  the  same  are  equivalent.  It  is 
therefore  consistent  to  represent  by  the  same  vector  all  couples 
whose  moments,  aspects,  and  senses  are  the  same.  Further,  it 
follows  from  art.  242  that  the  effect  of  a  couple  applied  to  a 
rigid  body  depends  only  upon  'its  moment,  aspect,  and  sense, 
which  are  therefore  the  essential  characteristics  of  a  couple. 
Hence  we  may  say  that  a  vector  completely  represents  a  couple, 
although  it  does  not  give  the  forces  or  arm  of  the  couple  nor  the 
position  of  its  plane. 

31.  Resolution  of  a  Force  into  a  Force  and  a  Couple. — 
Proposition. — ^A  force  may  be  resolved  into  a  force  acting 
through  any  arbitrarily  chosen  point  and  a  couple. 

Proof:  Let  F  (fig.  i6a)  denote  the  force  to  be  resolved,  and 
P  the  chosen  point,  a  distant  from  F.  Imagine  two  opposite 
forces  equal  and  parallel  to  F  introduced  at  P  (fig.  166).  Ob- 
viously the  three  forces  of  this  figure  are  equivalent  to  the  given 
force,  i.e.,  they  are  components  of  it.  But  the  three  forces  may 
be  grouped  into  a  force  at  P  and  a  couple. 


Observe  that  the  component  force  has  the  same  magnitude 
and  direction  as  the  given  force,  and  that  the  moment  of  the 
component  couple  is  the  same  as  that  of  the  given  foi^ce  about 
the  chosen  point. 


2  2  EQUIVALENCE   OF  FORCE  SYSTEMS.  [Chap.  II. 

Since  couples  whose  moments,  aspects,  and  senses  are  the 
same  are  equivalent,  the  force  F  (iig.  i6a)  is  equivalent  to  F 
of  Fig.  1 6c  and  the  couple  there  represented,  provided  that  its 
moment  equals  F-a. 

EXAMPLE. 

Resolve  the  8 -lb.  force  (fig.  6)  into  one  acting  at  B  and  a 
couple;  into  one  acting  at  C  and  a  couple. 

§  II.     CoLLiNEAR  Forces. 

32.  Composition.* — Let  F^,  F^,  F^,  etc.  (fig.  17),  denote  the, 
forces  to  be  compounded.     From  art.  21 
f-<i — ^-^- — )>    >    >    it    follows   that    the    resultant  of  those 
forces   of   the   system  having  the    same 
sense  is  equal  to  their  sum,  that  is, 


2   ^v /  -^i 

Fig.  17. 


the  resultant  of  F^,  F^,  F^,  etc.,  or  R\  =F^+F^-\-F^+  . . .  , 
and 

the  resultant  oi  F^,  F^,  F^,  etc.,  or  R" ,=F^-YF ^-^-Fq^-  .... 

Also,  the  resultant  of  the  system,  or  R,  equals  the  difference 
between  R'  and  R" .  Hence  if  R  be  given  sign,  positive  or  nega- 
tive according  as  it  acts  right  or  left, 

R=R'-R''  =  ^F,+F,+F,+  ...)-{F,+F,+F,+  ,,.) 
=  F,-F,+F,-F,+  .., 

The  action  line  of  R  is  of  course  the  same  as  that  of  the  given 
forces. 

§  III.     CoPLANAR  Concurrent  Non-parallel  Forces. 

33.  Graphical  Composition. — Let  AB,  BC,  CD,  and  DE 
(fig.  18)  be  the  forces  to  be  compounded.  By  the  triangle 
law   (art.  20),  compound   AB  and  BC   and   replace   them   by 

*  In  the  following  articles  on  composition  it  is  assumed  that  the 
force  systems  are  applied  to  rigid  bodies. 


§111.]       COPLANAR   CONCURRENT  NON-PARALLEL  FORCES. 


23 


their  resultant  AC;  compound  AC  and  CD  and  replace  them 
by  their  resultant  AD;  finally  compound  AD  and  DE  and  re- 
place them  by  their  resultant  AE.  By  successive  replacements, 
the  given  system  has  been  reduced  to  a  single  force,  AE,  which 
is  therefore  the  resultant  sought.     Notice  that  the  lines  AC,  AD, 


ac  and  ad  are  not  necessary  in  a  solution;  they  are  used  here 
only  for  demonstration  purposes. 

34.  Force  Polygon. — The  figure  formed  by  drawing  in  suc- 
cession lines  representing  the  magnitude  and  direction  of  the 
forces  of  any  system  is  called  a  force  polygon  for  those  forces, 
or  for  the  system.  In  fig.  18,  ABCDE  is  a  force  polygon  for 
the  given  system.  Several  force  polygons  can  be  drawn  for  any 
system,  one  for  each  possible  order  of  drawing  the  lines;  thus, 
for  the  system  compounded  in  the  preceding  article  twenty-four 
polygons  can  be  drawn,  no  two  alike. 

Proposition. — The  algebraic  sum  of  the  components  of  any 
system  of  coplanar  forces  along  any  line  equals  the  component 
along  that  line  of  a  force  whose  magnitude  and  direction  are 
represented  by  the  line  drawn  from  the  begin- 
ning to  the  end  of  the  polygon  for  the  system. 

Proof:  Let  ABCDE  (fig.  19)  be  the  polygon 
for  the  system.  The  components  of  the  forces 
along  the  line  A'C  are,  in  magnitude  and  direc- 
tion, A'B',  B'C\  C'D\  D'E\  and  that  of  a  force 
whose  magnitude  and  direction  are  AE  is  repre- 
sented by  A'E'.  From  the  figure,  it  is  plain 
that  A'E'  equals  the  algebraic  sum  of  A'B',  B'C\  C'D\  and 
D'E'. 


E'  B'  D' 

Fig.  190 


24 


EQUIVALENCE   OF  FORCE  SYSTEMS. 


[Chap.  II. 


35.  Rule  jor  Composition. — Draw  a  force  polygon  for  the 
forces;  then  a  line  from  the  beginrang  w  the  end  of  it.'  That 
line  represents  the  magnitude  and  direction  of  the  resultant; 
its  action  line  passes  through  the  common  point  of  the  action 
lines  of  the  given  forces. 

Examining  fig.  18,  it  will  be  seen  that  the  vector  sum  of  the 
given  forces  represents  the  magnitude  and  direction  of  the 
resultant. 

EXAMPLES. 

^i.  Determine  the  resultant  of  the  forces  represented  in 
fig.  6. 

2.  Solve  the  preceding  example,  taking  the  forces  in  a  differ- 
ent order  in  the  force  polygon. 

36.  Algebraic  Composition.  —  Let  fig.  20(a)  represent  the 
forces  to  be  compounded  and  the  body  to  which  they  are  applied. 
Resolve  each  force  at  the  origin  into  its  x-  and  ^-components, 
and  replace  it  by  them;  the  resulting  system  is  represented  in 

y 


(a) 


(b)  (c) 

Fig.  20. 


fig.  20(6).  Next  compound  all  the  ^-components  and  replace 
them  by  their  resultant,  ^Fx,  and  compound  the  ;y-components 
and  replace  them  by  their  resultant,  -TF^;  the  resulting  system 
is  represented  in  fig.  20(c).  Now  these  three  systems  are  equiva- 
lent and  have,  therefore,  the  same  resultant.  If  R  denotes  the 
resultant  and  d  its  direction  angle,  from  fig.  2o(<i)  it  is  plain 
that 

cos  d  =  IFx/R     and     sin  ^  =  JF^/R, 

and  the  action  line  of  R  contains  the  common  point  of  those  of 
the  given  forces. 


§IV.]       COPLANAR  NON-CONCURRENT  PARALLEL  FORCES. 


25 


Referring  to  the  figure,  or  to  prop.,  art.  34,  it  is  plain  that 
the   resolved    part   of  R    along   any   line  ^f^  p, 

equals  the  algebraic  sum  of  resolved  parts 
of  its  components  along  the  same  line,  that 
is, 

Rcos^  =  2'F;,. 

EXAMPLES. 

1.  Solve  ex.  i  of  the  preceding  article 
algebraically. 

2.  Let   Fj,  F2,  etc.,  fig.  21,   equal  8,  4,  /^'  ^^' 

6,  12,  7,  and  5  lbs.  respectively,  and  compute  their  resultant. 


1  \ 

[      / 

^x 

1 

§  IV.   CoPLANAR  Non-concurrent  Parallel  Forces. 

37.  Graphical    Composition.  —  Let  AB,  BC,  CD,  and    DE 
(fig.  22)  be  the  forces  to  be  compounded.     ABODE  is  a  force 


^ 

-^C7e~^-^ 

B* 

A 
D 

6 

' —    c 
c 

d 

Jf'o 


,-«■/ 


.(a) 


(b) 


Fig.  22. 

polygon  for  the  given   forces,  and  O  is  any  arbitrarily  chosen 

point. 

First,  resolve  AB  into  AO  and  OB, 

BC     "     50     -    OC, 

CL)     "     CO     "     OD, 
and  DF     "     Z)0    "    OE. 

Next  replace  each  given  force  by  its  components.      By  art.  24, 
the  action  lines  oi  AO  and  OB  may  intersect  anywhere  on  ah. 


It 

tt 

« 

"  CO 

"    OD 

tt 

tl 

tt 

''  DO 

'^    OE 

"    de. 


26  EQUIVALENCE  OF  FORCE  SYSTEMS.  [Chap.  IL 

It  is  therefore  possible  to  choose  the  action  Hnes  of  the  com- 
ponents so  that  those  of  OB  and  Bp,  OC  and  CO,  and  OD  and 
DO  coincide.  Having  so  taken  them,  it  is  plain  that  OB  and 
BO,  OC  and  CO,  OD  and  DO,  balance;  therefore  the  system 
of  components  reduces  to  ^O  and  OE.  Finally,  the  resultant 
of  AO  and  OE  is  AE  (art.  20),  which  is  also  the  resultant  of 
the  given  system. 

38.  Funicular  Polygon,  etc. — The  point  O  (fig.  22)  is  called 
the  pole  of  the  force  polygon.  Lines  OA,  OB,  OC,  etc.,  from 
the  pole  to  the  vert  exes  of  the  force  polygon  are  called  rays. 
Lines  oa,oh,oc,  etc.,  are  called  strings,  which,  considered  collect- 
ively, are  called  a  string,  or  funicular  polygon. 

If  the  notation  for  graphical  statics  (art.  11)  be  used,  the 
following  rules  for  drawing  the  funicular  polygon  will  be  found 
helpful,  but  the  beginner  should  not  depend  entirely  on  them. 

(a)  The  two  strings  intersecting  on  the  action  line  of  any 
force  are  parallel  to  the  rays  drawn  to  the  ends  of  the  vector 
corresponding  to  that  force;  or,  the  strings  intersecting  on  mn 
are  om  and  on. 

(b)  A  string  joining  points  in  the  action  lines  of  two  forces  is- 
parallel  to  the  ray  drawn  to  the  common  point  of  the  vectors 
corresponding  to  those  forces;  or,  the  string  joining  points  on 
Im  and  mn  is  parallel  to  OM. 

39.  Rule  for  Composition. — Draw  a  force  and  a  funicular 
polygon  for  the  forces ;  then  a  line  from  the  beginning  to  the  end 
of  the  force  polygon,  and  a  parallel  one  through  the  intersection 
of  the  first  and  last  strings  of  the  string  polygon.  The  first  line 
represents  the  magnitude  and  direction  of  the  resultant,  and 
the  second  its  action  line.  (The  first  and  last  strings  are  those 
corresponding  to  the  rays  drawn  to  the  beginning  and  end  of  the 
force  polygon.) 

It  is  plain  from  fig.  22  that  the  vector  sum  of  the  given 
forces  represents  the  magnitude  and  direction  of  the  resultant. 

EXAMPLES. 

Fig.  23  represents  a  board  upon  which  several  parallel  forces 
are  applied.  In  each  e?cample  below  determine  completely  the 
resultant. 


§IV.]  •     COP  LAN  A  R  NON-CONCURRENT  PARALLEL  FORCES.  27 

1.  The  magnitudes  of  F^,  F^,  etc.,  are  40,  10,  30,  20,  50,  and 
15  lbs.  respectively.  Solve  twice,  drawing  two  funicular  poly- 
gons, starting  them  at  different  points. 

2.  The  magnitudes  of  F^,  F^,  etc.,  are  20,  10,  30,  30,  10, 
and  o  lbs.  respectively.  Solve  twice,  drawing  two  funicular 
polygons  using  different  poles. 

40.  The  Resultant  when  the  Force  Polygon  Closes. — In  that 
case,  the  beginning  and  end  of  the  force  polygon  {A  and  E^ 


< 

8  ft.--— 

V 

1 

'     > 

■   •  1  ■•  M  1  ■ 

1 

Fig.  23. 

fig.  22)  coincide;   hence  the  first  and  last  strings  are  parallel. 
The  forces  corresponding  to  these  strings,  and  to  which   the 
given  system  is  equivalent,  are  equal  and  opposite  (AO  and  OE] 
hence  they  constitute  a  couple  and  can  not  be  compounded. 

If  the  strings  ao  and  oe  should  happen  to  coincide,  then  the 
forces  AO  and  OE  would  balance;  and  their  resultant,  and  hence 
that  of  the  given  system,  is  zero.     Therefore 

A  funicular  polygon  for  a  system  whose  force  polygon 
closes  may  be  open  or  closed;  if  open,  the  resultant  is  a 
couple,  and  if  closed,  the  resultant  is  zero. 

It  is  worth  noting  that  in  the  second  case,  force  and  funicular 
polygons  closed,  each  segment  of  the  funicular  polygon  is  the 
action  line  of  two  forces. 

EXAMPLES. 

I.  The  magnitudes  of  F^,  F^,  etc.  (fig.  23),  are  20,  55,  o,  15, 
10,  and  30  lbs.  respectively.     Determine  their  resultant. 

Solution:  The  force  polygon  is  ABCDEF  (fig.  24),  and 
since  it  is  closed,  the  resultant  is  a  couple.  If  we  regard  A  as 
the  beginning  of  the  force  polygon,  the  system  compounds  into 
two  forces  AO  and  OF  whose  action  lines  are  ao  and  of  respec- 
tively. They  constitute  the  resultant  couple;  its  arm  is  the 
dotted  line  of  the  space  diagram.     The  arm  scales  2.58  ft.  and 


28 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


the  forces  67.9  lbs.,  hence  the  moment  of  the  couple  is  175.2 
ft.-lbs.     The  sense  is  clockwise. 

2.  Solve  the  preceding  example,  taking  the  forces  in  a 
different  order  in  the  force  polygon. 

41.  The  Principle  of  Moments. — The  algebraic  sum  of  the 
moments  of  any  number  of  coplanar  parallel  forces  with  respect 


o 


b   b 


b.. 


F 
A  — 


■^«-4 


V\, 


~1 d 

o  ~ 

\    .X -- 

— —     o 

Scale:   lin.=  4ft. 

^._    (a) 


■~~-~r^^^  o 


^^^        Scale:   lin.=  40U)8. 
--^^  (b) 

Fig.  24. 

to  any  origin  in  their  plane  equals  the  moment  of  their  resultant 
with  respect  to  the  same  origin. 

Proof:  Let  the  system  be  that  represented  in  fig.  22.     Re- 
membering 

that  AO  and  OB  are  concurrent  components  oi  AB, 
"    BO    ♦'    OC    "  "  "  "  BC, etc., 

and  recalling  also  Varignon's  theorem,  we  may  write 

moment  of  A B  =  moment  of  ylO  +  nioment  of  OB; 
moment  of  -BC  =  moment  of  BO  +  moment  of  OC; 
moment  of  CD  =  moment  of  CO  +  moment  of  OD; 
moment  of  Z)£= moment  of  DO + moment  of  OE. 

Therefore 

2'(moments  of  AB,  BC,  CD,  and  DE) 

=  ^(moments  of  AO,  OB,  BO,  OC,  CO,  OD,  DO,  and  OE). 


§IV.]        COPLAN/lR  NCN-CCNCURRENT  PARALLEL  FORCES.  29 

But  the  moments  of  OB  and  BO,  of  OC  and  CO,  etc.,  are  equal 

and  opposite,  hence 

^(moments  of  A5,  ^C,  CZ),  and  Z)£) 

=  moment  of  ^O  +  moment  of  OE, 
If  the  resultant  is  a  single  force, 

moment  oi  A0  +  moment  of  OE  =  moment  oi  AE\ 
if  the  resultant  is  a  couple, 

moment  of  AO  +  0-E  =  moment  of  the  couple  (see  art.  29). 

Therefore,  in  either  case,  the  sum  of  the  moments  oi  AB,  BC^ 
CD,  and  DE  =  th.e  moment  of  their  resultant.  Proof  may 
readily  be  extended  to  a  system  of  more  than  four  forces. 

42.  Algebraic  Composition. — I.  //  the  algebraic  sum  of  the 
forces  is  not  zero,  the  force  polygon  does  not  close  (art.  34); 
hence  the  resultant  is  a  force  (art.  37).  li  F^,  F^,  etc.,  denote 
the  forces  and  R  their  resultant, 

R  =  i'F, 

and  the  sense  of  R  is  given  by  the  sign  of  IF. 

The  action  line  may  be  determined  by  means  of  the  prin- 
ciple of  moments;  thus,  if  IMq  denotes  the  algebraic  sum  of 
the  moments  of  the  forces  with  respect  to  any  origin  O,  and 
a  the  corresponding  arm  of  their  resultant  Rt 

2'Mo  =  Ra,     or    a  =  i'Mo/R. 

iThe  resultant  must  act  on  that  side  of  O  which  will  make  the 
sign  of  its  moment  the  same  as  that  of  IMq. 

II.  //  the  algebraic  sum  of  the  forces  is  zero,  the  force  poly- 
gon closes  (art.  34);  hence  the  resultant  is  a  couple  (art.  40). 
According  to  the  principle  of  moments,  the  moment  of  this 
couple  equals  the  algebraic  sum  of  the  moments  of  the  given 
forces  about  any  point. 

EXAMPLES. 

1.  Solve  ex.  i,  art.  39,  three  times,  using  each  time  a  new 
origin  of  moments. 

2.  Solve  ex.  2,  art.  39-,  twice,  using  different  origins  of 
moments. 

3.  Solve  ex.  i,  art.  40. 


30 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


43.  Two  Unequal  Parallel  Forages. — This  is  a  common  case  to 
which  the  following  special  methods  may  be  applied.  Let  P 
and  Q  denote  the  forces,  P  the  larger,  and  R  their  resultant. 


^  o 

Jr       p     ^^-^ 
'(a)  (b) 

Fig.  25. 
When  P  and  Q  are  alike  in  sense  (see  fig.  25a), 

R  =  P  +  Q. 
the  sense  of  R  is  the  same  as  that  of  the  given  forces, 
X  =  Qc/R     and     y  =  Pc/R . 
When  P  and  Q  are  unlike  in  sense  (see  fig.  256), 
R  =  P-Q, 
the  sense  of  R  is  the  same  as  that  of  the  larger  force, 
x  =  Qc/R     and    y  =  Pc/R.* 

The  student  should  prove  the  expressions  for  x  and  y  and  show 
that  R  is  correctly  represented  in  the  figure,  i.e.,  that  its  action 
line  is  between  the  forces  in  (a) ,  but  beyond  the  larger  force 
in  (6). 

The  following  relations  are  sometimes  convenient: 

From  either  figure  Px  =  Qy,  therefore 

PAC  =  QBC,     or 

AC/BC  =  Q/P; 

hence  the  action  line  of  the  resultant  of  two  parallel  forces 
divides  any  secant  intersecting  their  action  lines  into  two  seg- 

*  It  is  plain  from  these  equations  that  the  smaller  R  is  (P  and  Q 
nearly  equal)  the  greater  x  and  y  are.  As  P  and  Q  approach  equality, 
they  become  more  nearly  equivalent  to  a  couple;  also  R  approaches  zero, 
and  its  arm  with  respect  to  any  origin  in  the  body  to  which  P  and  Q 
axe  applied  approaches  00  .  Hence  we  arrive  at  the  conception  that  a  couple 
is  equivalent  to  a  force  of  zero  magnitude  with  an  infinite  arm. 


v.]    COPLANAR  NON-CONCURRENT  NON-PARALLEL   FORCES,       3^ 


ments  which  are  inversely  proportional  to  the  magnitudes  of 
those  forces.     Again, 

P/BC  =  Q/AC=^{P  +  Q)/{BC-{-AC)  =  {P-Q)/{BC-AC), 

or 

P/BC^Q/AC^R/AB\ 

'  hence  the  forces  P,  Q,  and  R  are  proportional  to  the  distances 
between  the  other  two. 


\ 


EXAMPLES. 


1.  Determine  the  resultant  of  F^  and  F^  of  ex.  i,  art.  39. 

2.  Determine  the  resultant  of  Fg  and  F^, 


§  V.     CoPLANAR  Non-concurrent  Non-parallel  Forces. 

44.  Graphical  Composition. — First  method.  This  consists 
in  compounding  two  of  the  forces,  then  their  resultant  and  a 
third  force,  that  resultant  and  a  fourth,  etc.,  until  the  simplest 
equivalent  system  has  been  found.     Thus,  let  AB,  BC^  CD, 

\ 


I 


Fig.  26. 

and  DE  (fig.  26)  be  the  forces  of  such  a  system.     According  to 
art.  20, 

the  resultant  oi  AB  and  BC  is  AC, 
;  "        **  "  AC     ''     CD   "  AD, 

and  "        ''  ''  AD     *'    DE  "  AE, 

Therefore  AE  is  the  resultant  sought. 

It  may  happen  that  some  of  the  intersections  in  the  space 
diagram  which  are  necessary  do  not  fall  within  convenient 
limits.  This  is  apt  to  occur  when  the  action  lines  of  the  given 
forces  are  nearly  parallel.     In  such  cases  it  is  practically  im- 


32 


EQUIVALENCE   OF  FORCE  SYSTEMS. 


rCHAP.  II, 


possible  to  determine  the  action  line  of  the  resultant,  and  the 
following  method  should  be  employed. 

Second  method.     This  is  the  same  as  that  for  the  composi- 
tion of  coplanar  non-concurrent   parallel  forces   explained   in 


-^V 


. — — '"     -^^"Z 


'     (a)  (b)         E 

Fig.  27. 

arts.  37,  T^'^,  and  39,  which  the  student  should  read  in  connection 
with  fig.  27. 

EXAMPLES. 

I.  Let  Fj,  F2,  etc.,  in  fig.  28  equal  8,  4,  6,  7,  12,  and  5  lbs. 
respectively,  and  determine  their  result- 
ant.* 

2.  Solve  ex.  i,  drawing  a  second  fu- 
nicular polygon  in  the  same  space  diagram 
beginning  it  at  some  other  point. 

3.  Solve  ex.  i,  choosing  a  new  pole 
but  employing  the  same  space  diagram. 

45.   The  Resultant  when  the  Force  Poly- 
gon Closes. — Just  as  in  the  case  of  parallel 
forces,  the  resultant  is  a  couple.     In  fact 
the  explanation  in  art.  40  may  be  read  in  connection  with  fig.  27. 

46.  The  Principle  of  Moments. — The  algebraic  sum  of  the 
moments  of  any  number  of  coplanar  forces  with  respect  to  any 
origin  equals  the  moment  of  their  resultant  with  respect  to  the 
same  origin. 

The  proof  given  in  art.  41,  read  in  connecti9n  with  fig.  27, 
applies  to  this  proposition. 

47.  Algebraic  Composition. — I.  7/  the  algebraic  sum  of  the 
X  and   y-components   of   the  forces  are  not  both  zero,  the    force 

*  Use  scales  not  less  than  i  in.  =2  ft.  and  i  in.  =4  lbs. 


Fig.  28. 


§  v.]     COP  LAN  A  R  NON-CONCURRENT  NON-PARALLEL  FORCES.       33 


polygon  for  the  system  does  not  close  (art.  34);    hence  the  re- 
sultant is  a  force  (art.  44).     Since 

Rx  =  IFx     and     Ry  =  IFy, 

R=(jF;'+iF;y, 

sin  d  =  iTy/R     and     cos  0  =  JF^/R, 
6  being  the  direction  angle  of  R  measured  from  the  x  axis. 

The  action  line  of  the  resultant  may  be  found  by  the  principle 
of  moments.  If  IMq  denotes  the  algebraic  sum  of  the  moments 
of  the  forces  with  respect  to  any  origin  0 ,  and  a  the  correspond- 
ing arm  of  R, 

im,  =  Rai,     or     a  =  i'M,/R. 
The  resultant  must  act  on  that  side  of  O  which  will  make  the 
sign  of  its  moment  the  same  as  that  of  2Mq. 

II.  //  the  algebraic  sums  of  the  x  and  y-components  equal 
zero,  the  force  polygon  for  the  system  closes  (art.  34);  hence 
the  resultant  is  a  couple  (art.  45).  According  to  the  principle 
of  moments,  the  moment  of  the  couple  equals  the  algebraic  sum 
of  the  moments  of  the  given  forces  about  any  origin. 

For  a  system  of  couples,  IF^  and  IFy  equal  zero;  hence 
their  resultant  is  a  couple,  and  its  moment  equals  the  algebraic 
sum  of  the  moments  of  the  given  couples. 

EXAMPLES. 

I.  Solve  ex.  i,  art.  44,  algebraically. 

Solution:  It  will  be  convenient  to  tabulate  the  computation. 


F 

a 

a 

^z 

■   ^v 

F.a 

8  lbs. 

4  " 

6  " 

7  '• 
12  " 

5  " 

0 

63    25 
90° 

36°  52' 
14°    2' 

2.00  ft. 
1. 41   " 
1.79  '' 
4.00  " 
3.20  " 
2.91  «* 

+8.00 

+2.83 

-2.68 

0.00 

-9.60 

+  4.85 

0 

+2.83 

+  5-3^ 
-7.00 
-7.20 
—  1. 21 

—  16.00 

-  5-64 
-10.74 
+  28.00 
-38.40 

+  14.55 

+3.40 

-7.22 

-28.23 

In  the  first  column  the  force  magnitudes  are  recorded,  in  the 
second  the  acute  angle  between  each  force  and  the  x  axis  (taken 
horizontal),  and  in  the  third  the  arms  of  the  force  with  respect 
to  the  selected  origin  (centre  of  the  board);  then  in  the  fourth, 
fifth,  and  sixth  columns  the  computed  values  of  the  ^c-conipo- 


34 


EQUiyALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II, 


The    algebraic 
-28.23   ft.-lbs. 


nents,  the  ^^-components,  and  the  moments, 
sums  of  these  are  3.40  lbs.,  —7.22  lbs.,  and 
respectively.     Hence 

(i)  i^  =  \/3'.4o  +  7l22  =  7.98  1bs.; 

(2)  the  sense  of  R  is  downward  to  the  right,  the  angle  with  the 

^-axis  being 

sin~^  7.22/7.98  =  64°  47'; 

(3)  the  action  line  of  R  is  to  the  right  of  the  centre  of  the  board 

a  distance  28.2^/7.98  =  3.54  ft. 

2.  Solve  the  preceding  example,  choosing 
a  different  origin  of  moments. 

3.  Let  F^,  F2,  etc.  (fig.  28),  be  20,  14.14, 
22.36,  15,  25,  and  o  lbs.  respectively.  Com- 
pute their  resultant. 

4.  Let  Fi,  F2,  and  F^  (Fig.  29)  be  10,  20, 
and  30  lbs.  respectively.     Determine  the  re- 

FiG.  29.  sultant.  the  square  being  4  x4  ft. 

48.  Reduction  of  a  System  to  a  Force  and  a  Couple. — 
Proposition. —  Any  system  of  forces  can  be  compounded  into  a 
force  acting  through  any  arbitrarily  chosen  point  and  a  couple. 

Proof:  *  Each  force  of  the  given  system  may  be  replaced  by 
a  force  acting  through  the  chosen  point  and  a  couple  (art.  31). 
Suppose  such  a  replacement  made  for  each  force;  the  resulting 
system  consists  of  a  concurrent  one  and  a  system  of  couples. 
But  the  resultant  of  the  concurrent  forces  is  a  single  force  acting 
through  the  chosen  point  (art.  S3  01*  36),  and  the  resultant  of  the 
couples  is  a  single  couple  (art.  47). 

Computation  of  the  force  and  the  couple.  Let  F  (fig.  30) 
be  one  of  the  forces  of  the.  given  system 
and  O  the  chosen  point.  The  components 
of  F  are  F'  (applied  at  O)  and  the  couple 
FF^\  Observe  now  that  the  component 
of  F'  along  any  line  is  the  same  in  magni- 
tude and  sense  as  that  of  F  along  the  same 

line  and  that  the  moment  of  the  couple,  FF^\  is  the  same  as  that     '. 

__^ ] ' I 

*  This  proof  is  for  a  coplanar  system,  the  chosen  point  being  in  the      I 
plane  of  the  forces.     Proof  for  the  general  case  is  given  in  art.  55.  \ 


Fig.  30. 


iJNIVERSITY 

OF 


§  VI.]  NON-COPLANAR  CONCURRENT  FORCES.  35 

of  F  about  O.  Hence  the  algebraic  sum  of  the  components  of  all 
the  forces  of  the  concurrent  system  along  any  line  is  the  same 
as  that  of  the  components  of  the  given  forces  along  that  line,  and 
the  sum  of  the  moments  of  the  couples  equals  the  sum  of  the 
moments  of  the  given  forces  about  0. 

If  7^0  denotes  the  resultant  of  the  concurrent  system^  and  C 
that  of  the  couples, 

sin  6  =  2'F^/R,     cos  6  =  IF^/R 
0  being  the  direction  angle  of  the  action  line  of  R,  and  C  =  i'Fa. 

EXAMPLE. 

Let  Fi,  F^,  etc.  (fig.  28)  equal  14,  10,  3,  9,  18,  and  11  lbs. 
Reduce  them  to  a  force  at  the  centre  and  a  couple. 

§  VI.       NON-COPLANAR    CONCURRENT    FoRCES. 

49.   Graphical    Composition.  —  Imagine     a     force     polygon 

ABC  ,  .  .  N  drawn  for  the  forces  to  be  compounded;  it  is  not 
of  course  a  plane  one.  According  to  the  triangle  of  forces,  the 
resultant  of  the  first  two  forces,  R',  is  represented  in  magnitude 
and  direction  by  AC.  Likewise  the  resultant  of  R'  and  the 
third  force,  i.e.,  the  resultant  of  the  first  three  forces,  is  repre- 
sented in  magnitude  and  direction  by  AD,  etc.  Finally,  the 
resultant  of  all  the  given  forces  is  represented  by  the  line  AN^ 
joining  the  beginning  and  the  end  of  the  force  polygon. 

Hence  the  magnitude  and  direction  of  the  resultant  is 
represented  by  the  vector  sum  of  the  given  forces,  or,  other- 
wise stated,  by  the  line  drawn  from  the  beginning  to  the  end 
of  the  force  polygon  for  the  system. 
The  action  line  of  the  resultant,  of  course,  passes  through  the 
common  point  of  the  action  lines  of  the  given  forces. 

Since  the  force  polygon  is  not  plane,  it  is  practically  neces- 
sary to  represent  it  by  projections.  The  line  representing  the 
resultant  may  then  be  determined  by  its  projections,  in  the 
projections  of  the  polygon. 

EXAMPLES. 

I.  Suppose  three  forces  F^,  F^,  and  F^  to  act  at  a  point  0  as 
shown    in    fig.  31(a).     The   horizontal  and  vertical  projections 


36 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  U. 


of  O  are  marked  O'  and  O"  respectively,  those  of  the  vector 
representing  F^  are  marked  F/  and^F/',  etc.* 

Solution:  At  any  point  A  we  begin  to  construct  the  polygon 
for  the  ioTCQS—A'B'CD'  is  its  horizontal  and  A"B"C"D"  is  its 
vertical  projection.     Hence  the  line  AD  represents  the  magni- 


FlG.  31. 

tude  and  direction  of  the  resultant;  the  action  line  passes 
through  O. 

If  the  horizontal  and  vertical  projections  of  the  vectors 
representing  the  given  forces  be  regarded  as  force  systems,  then 
obviously  A'B'C'D'  and  A"B"C"D''  are  the  force  polygons  for 
those  systems  respectively. 

2.  Imagine  the  forces  of  fig.  38(^7)  to  act  in  the  directions 
indicated  but  through  the  corner  diagonally  opposite  O,  and 
determine  their  resultant. 

50.  Algebraic  Composition. — Let  F',  F",  F"',  etc.,  denote 
the  forces  to  be  compounded.  At  the  common  point  of 
their  lines  of  action,  resolve  each  force  into  three  rectangular 
{x,  y,  and  z)  components  and  replace  the  forces  by  them.  Next, 
compound  separately  all  the  x,  y,  and  ^-components  and  re- 
place them  by  their  resultants,  IF^,  ^Fy,  and  IFg.  Now  these 
three  systems,  the  given  one,  the  system  of  components,  and 
the  three  forces,  IF^;,  IFy,  and  IFg,  are  equivalent;  they  have, 
therefore,  the  same  resultant.     If  R  denotes  the  resultant,  and 


*  All  horizontal  and  vertical  projections  are  marked  by  primes  and 
double  primes  respectively. 


5  VII.] 


NON-COPLANAR  PARALLEL  FORCES. 


37 


» 


6^,  6^,  and  6^  the  direction  angles  of  its  action  line,  from  the  third 
system  it  is  plain  that 

(i)  r=(jf;'+if"/+^/)^ 

(2)  cos^i  =  i'F;,/R,    cos02  =  IFy/R,     cosd^  =  I¥,/R, 

(3)  the  action  line  passes  through  the  common  point  O. 

EXAMPLE. 

Solve  ex.  2,  art.  49,  by  the  method  of  this  article. 

§VII.       NON-COPLANAR    PARALLEL    FORCES. 

51.  Graphical  Composition. — The  vectors  F^  and  F2  (fig.  32) 
represent  two  parallel  forces,  F/ 
and  F^  their  projections  on  the  xz 
plane,  and  F/'  and  F/'  those  on  the 
yz  plane.  Let  R  represent  the  re- 
sultant of  the  two  forces,  and  R'  and 
i?"  the  projections  of  the  vector  R 
upon  the  plane  xz  and  yz  respect- 
ively. 


If  F, 


and    F2' 


be   regarded    as 


forces,  F'  represents   their  resultant 
for 


FJF,  =CB/CA  -=C'B'IC'A'  -F^fF.!, 

i.e.,  R'  divides  the  line  A'B'  into  segments  inversely  propor- 
tional to  F/  and  F/;  and  obviously  F'=F/+F2'.  Similarly, 
if.F/'  and  F^'  be  regarded  as  forces,  R'!_  represents  their  re- 
sultant.    To  find  the  resultant,  then,  of  two  parallel  forces, 

Project  the  vectors  representing  them  upon  two  planes 
parallel  to  the  forces,  and  find  the  resultants  of  these  pro- 
jections regarded  as  forces.  These  resultants  are  projec- 
tions of  the  resultant  sought. 

This  method  may  obviously  be  extended  to  the  composition  of 
more  than  two  forces. 

EXAMPLE. 

Suppose  parallel  forces  of  14,  12,  16,  and  8  lbs.  applied  to  a 
body  at  points  whose  x  and  y  coordinates  are  respectively  (2,4), 


38 


EQUIVALENCE  OF  FORCE  SYSTEMS. 


[Chap.  IL 


(3.  5).  (4»  2),  and  (6,  3),  all  in  feet,  and  suppose  that  the  third 
force  acts  in  the  negative  and  the  'others  in  the  positive  direc- 
tion. 

Solution:  The  vectors  representing  the  forces  are  projected 


-IP 

■"t'l(3,5) 


(2,i)] 


o^;;:: 


Fig.  33. 


on  the  yz  and  zx  planes;  thus  a'5'  and  a"5"  (fig.  33)  are  the 
projections  of  the  14-lb.  force,  6'c'  and  ^"c"  those  of  the  12 -lb, 
force,  etc.  Each  projected  system  may  be  compounded  by 
the  method  of  art.  37,  one  force  polygon,  as  ABODE,  sufficing 
for  the  two  systems.  The  funicular  polygon  in  the  xz  plane 
determines  a^e',  and  that  in  the  yz  plane  determines  a"^".  The 
action  line  of  the  resultant  sought  then  passes  through  the  point 
P  in  the  xy  plane.  The  magnitude  and  sense  of  the  resultant 
are  given  by  ^-E. 

52.  The  Resultant  when  the  Force  Polygon  Closes  is  a 
couple.  For,  obviously,  the  resultant  of  all  the  forces  of  the 
system  but  one  is  equal  and  opposite  to  that  one;  hence  that 
resultant  and  the  last  force  constitute  the  resultant  couple. 
The  forces  and  arm  of  this  couple  depend  on  which  force  of 
the  system  is  omitted;  accordingly,  the  system  may  be  reduced 


.§VII.] 


NON-COPLANAR  PARALLEL  FORCES. 


39 


by  this  method  to  as  many  different  couples  as  there  are  forces 
in  the  system,  but  they  are  equivalent  to  each  other. 

Let  the  system  consist  of  F^,  F^,  F^,  etc.,  and  let  R  denote 
the  resultant  of  F^,  F^,  etc. ;  then  the  resultant  couple  consists 
of  F^  and  R.  Imagine  the  vectors  representing  the  forces  to  be 
projected  on  two  planes  and,  as  above,  denote  the  projections 
on  one  plane  by  F/,  F^ ,  F^ ,  .  .  .  and  R',  and  those  on  the 
other  by  F/',  F/',  F3",  .  .  .  and  R''.  Evidently,  the  resultants 
of  the  systems  F/,  F/,  F/,  etc.,  and  F/',  Fj",  F3",  etc.,  are 
couples,  and  the  resultant  of  the  first  is  F/,  R',  and  that,  of 
the  second  is  F/',  R'\ 

It  might  of  course  happen  that  F^  and  R  coincide.  In  that 
case  the  resultant  of  the  given  system  would  be  zero ;  and  since 
F/  and  R'  and  F^"  and  i?"  would  also  coincide,  the  resultants 
of  the  systems  F/,  Fj',  F3',  etc.,  and  F/',  Fj",  F3",  etc., 
would  be  zero,  and  the  funicular  polygons  for  those  systems  would 
close  (art.  40). 

EXAMPLE. 

Solve  exs.  i  and  2,  art.  54,  graphically. 

53.  The  Principle  of  Moments. — The  algebraic  sum  of  the 
moments  of  any  number  of  parallel  forces  v/ith  respect  to  a  line 
equals  the  moment  of  their  resultant  with  respect  to  that  line. 

Proof:  Let  F^  and  Fg  (fig.  34)  be  two  forces  of  the  system, 


Fig.  34. 
R'  their  resultant,  and  OP  the  moment  axis.     OP  is  regarded 
as  a  s  coordinate  axis,  and  the  y  axis  is  so  taken  that  the  yz 
plane  is  parallel  to  the  forces;    A,  B,  and  C  are  the  points 
where  F^,  Fg,  and    R'  respectively  pierce  the  xz  plane;    and 


40  EQUIVALENCE  OF  FORCE  SYSTEMS.  [Chap.  II-     ^ 

i 

Aa  and  Aa'  represent  the  components  of  F^  parallel  and  per-  \ 

pendicular  respectively  to  the  moment  axis,  Bh  and  Bh'  those  i 

of  Fj,  and  Cc  and  Cc'  those   of  R\     The  moments  of  the  three  '> 
forces  are  respectively 

(F,  sina:)AA',     {F,sma)BB\     and     (R' sin  a)CC^.  \ 

1 
Evidently,  the  perpendicular  component  of  R  is  the  resultant  ] 
of  the  perpendicular  components  of  Fj  and  Fj.  Then,  accord-  | 
ing  to  the  principle  of  moments  for  coplanar  forces,  j 


(R'  sin  a)OC  =  (F^  sin  a)OA  +  (F^  sin  a)OB. 

This  equation  multiplied  through  by  sin  (FOB)  becomes 

(F'  sin  a)CC'  =  (F,  sin  a)AA'  +  {F^  sin  a)BB'. 

That  is,  the  moment  of  F'  equals  the  sum  of  the  moments  of 
Fi  and  F^. 

If  Fj  and  Fg  have  unlike  senses,  a  slight  change  in  the  proof 
is  necessary,  which  the  student  can  make.  The  extension  of 
the  proof  to  more  than  two  forces  is  quite  evident. 

For  the  exceptional  case  in  which  the  resultant  is  a  couple, 
the  proposition  still  holds  if  we  define  the  moment  of  a  couple 
with  respect  to  a  line  to  be  the  algebraic  sum  of  the  moments  of 
its  forces  with  respect  to  that  line. 

54.  Algebraic  Composition. — I.  //  tAe  algebraic  sum  of  the 
forces  is  not  zero,  the  force  polygon  for  the  system  does  not 
close;  hence  the  resultant  is  a  force  (art.  51).  If  F^,  F2,  etc., 
•denote  the  forces  and  R  their  resultant, 

the  sense  of  R  being  given  by  the  sign  of  JF. 

The  action  line  may  be  determined  by  the  principle  of 
moments;  thus,  if  ^M^  and  ^My  denote  the  algebraic  sums  of 
the  moments  of  the  forces  with  respect  to  two  axes,  x  and  y, 
which  are  perpendicular  to  the  forces,  and  ax  and  ay  the  corre- 
sponding arms  of  the  resultant,  R, 

IMx  =  Rax     and     IMy  =  Ray, 
whence  ax  =  i'M^/R,     a.y  =  I'NLy/R. 


NON-COPLANAR,  NON-CONCURRENT,  NON-PARALLEL  FORCES.     41 

The  resultant  must  act  on  such  sides  of  the  x  and  y  axis  that 
the  signs  of  its  moments  are  the  same  as  those  of  IM^  and  ^My 
respectively. 

II.  //  the  algebraic  sum  of  the  forces  is  zero,  the  force  poly- 
gon closes;  hence  the  resultant  is  a  couple  (art.  52).  As  ex- 
plained in  the  article  referred  to,  the  resultant  couple  is  not 
determinate.  A  resultant  couple  can  be  readily  found  by  com- 
puting the  resultant  of  all  the  forces  of  the  system  but  one; 
this  resultant  force  and  the  omitted  force  constitute  a  couple, 
the  resultant  of  the  system. 

EXAMPLES. 

1.  Compute  the  resultant  of  five  forces  of  15,  12,  20,  16,  and 
21  lbs.,  the  first  three  acting  in  the  positive  z  direction  and  the 
last  two  in  the  negative,  the  coordinates  of  the  points  in  which 
they  pierce  the  xy  plane  being  respectively  (2,  3),  (4,  —2), 
(2,  4),  (3,  -I),  and  (o,  o). 

2.  Include  a  force  of  10  lbs.  acting  in  the  negative  z  direction 
at  a  point  whose  coordinates  are  (  —  8,  10),  and  compound. 

§  viii.     non-coplanar,  non-concurrent,  non-parallel 

Forces. 

55.  The  Resultant. — Proposition  I.  A  system  of  non- 
coplanar,  non-concurrent,  non-parallel  forces  raay  be  com- 
pounded into  two  forces,  the  action  line  of  one  being  in,  and 
that  of  the  other  normal  to,  any  arbitrarily  selected  plane. 

Proof:  Let  the  action  lines  of  the  forces  of  the  system  be 
extended  till  they  pierce  the  selected  plane.*  This  will  be 
referred  to  as  the  plane  7:.  At  each  of  these  points  resolve  the 
corresponding  force  into  two  components,  one  in  the  plane  and 
the  other  normal  to  it.  The  given  system  may  be  regarded  as 
replaced  by  two,  a  coplanar  system^  (the  components  in  tc),  and 
a  parallel  system  (the  components  normal  to  tt).  In  general, 
each  of  these  systems  compounds  into  a  single  force,  but  aither 

*  A  force  whose  action  line  is  parallel  to  the  plane  should  be  replaced 
by  its  equivalent,  a  force  in  the  plane  and  a  couple  whose  forces  are 
perpendicular  to  it.      (Art.  31.) 


4^                          EQUIVALENCE  OF  FORCE  SYSTEMS.              [Chap.  II.  ; 

-  .-i 

or  both  of  them  may  compound  into  a  couple  (arts.  45  and  51).  \ 

But,  as  shown  in  art.  42,  a  couple  <nay  be  regarded  as  "a  zero  i 

force  with  an  infinite  arm,"  and  with  this  understanding  the  \ 
proposition  holds  in  all  cases. 

Since  the  two    forces    do    not    usually   intersect,  it   is  not  j 

possible,  in  general,  to  compound  them  into  one  force.     They  ] 

may  therefore  be   properly  called   a   resultant    of  the    system.  \ 

For  different  planes  71  we  arrive  at  different  pairs  of  resultant  ^ 

forces,  but  since  they  are  equivalent  to  the  same  system  they  i 

are  equivalent  to  each  other.     We  will  denote  the  resultant  \ 

forces  in  and  normal  to  the  plane  by  Rt  and  Rn  respectively.  i 

Proposition  II.     A  system  of  non-coplanar,  non-concurrent,  \ 

non-parallel  forces  may  be   compounded   into   a   force   acting  \ 

through  any  arbitrarily  selected  point  and  a  couple.  j 

Proof:  Each  force  of  the  given  system  may  be  replaced  by  \ 

a  force  acting  through  the  chosen  point  and  a  couple  (art.  31).  i 

Suppose  such  a  replacement  made  for  each  force;  the  result-  ^ 

ing  system  consists  of  a  concurrent   system  of  forces   and   a  \ 

system  of  couples.     But  the  resultant  of  the  concurrent  forces  \ 

is  a  single  force  acting  through  the  chosen  point  (arts.  49  or  I 

50),  and  the  resultant   of   the   couples  is  a  single  couple  (art.  \ 

60).      We    will   denote    this    force    and    couple   by  R   and  C  ■ 

respectively.'  i 

In  general,  R  and  C  may  be  compounded  into  two  non-  j 

parallel  unequal  forces.     For,  C  may  be  replaced  by  an  equiva-  i 

lent  couple  one  of  whose  forces  intersects  R  (see  art.  59),  and  j 

that  force  and  R  may  be  compounded  into  a  force;  since  this  i 

last  force  will  not  be  in  the  same  plane  with  the  other  force  of  i 

the  couple  these  two  forces  cannot  be  compounded,  and  they  ; 

may  properly  be  called  a  resultant  of  the  system.  | 

If  the  plane  of  C  happens  to  be  parallel  to  R,  C  and  R  \ 

may  be  compounded  into  a  single  force.      For,  C  may  be  re-  . 

placed   by  an   equivalent  couple    whose  plane  coincides  with  : 

R\    and,  the  forces  of  that  couple  and  R  being  coplanar,  their  \ 

resultant  is  a  single  force.     That  force  is  the  resultant  of  the  \ 

system.  ! 

In  the  following  three  articles,  methods  are  explained  for  : 

determining  the  resultants  above  discussed.  ; 


NON-COP L/iN A R,  NON-CONCURRENT,  NON-PARALLEL  FORCES.     43 

56.  Graphical  Composition. — The  graphical  method  can  be 
employed  to  determine  Rt  and  Rn,  the  method  of  art.  44  for  Rt. 
and  that  of  art.  51  for  Rn.  Some  elementary  principles  of 
descriptive  geometry  are  employed  for  determining  the  inter- 
sections of  the  lines  of  action  of  the  forces  of  the  given  system 
with  the  plane  tz.  The  method  will  be  illustrated  by  means 
of  an 


EXAMPLE. 


Let  there  be  three  forces  in  the  system,  the  horizontal  pro- 
jections of  the  vectors  representing  them  being  F/,  F^,  and  F3' 
(fig.  35),  and  the  vertical  projections  F/',  F/',  and  F3". 


'//'"'»'" '°'^^         / 

Fig.  35. 

The  plane  it  is  taken  to  coincide  with  the  xy  plane,  the 
forces  piercing  it  in  the  points  P^,P^,  and  P^  respectively.  The 
components  of  the  given  forces  in  the  plane  tz  are  represented 
by  the  vectors  F/',  F^' ,  and  F3",  while  the  components  normal 
to  that  plane  act  through  the  points  P^,  P.^,  and  P3,  and  their 
magnitudes  are  represented  by  the  projections  of  F/,  F^ ,  and 
Fg'  upon  the  z  axis. 

The   resultant   of    the    coplanar   system   is    represented   in 


44  EQUIVALENCE   OF  FORCE  SYSTEMS.  [Chap.  11. 

magnitude  and  direction  by  A'D\  A'B'CD'  being  a  force  poly- 
gon for  the  system,  and  its  action  iine  is  marked  Rt^ 

The  resultant  of  the  normal  system  is  represented  in  magni- 
tude and  direction  hy  AD,  ABCD  being  a  force  polygon  for 
the  system.  The  funicular  polygon  for  the  projection  of  the 
normal  system  on  the  xz  plane  is  oa' ,  oh',  oc' ,  and  od\  and  Rn 
is  the  projection  of  Rn  on  that  plane.  The  funicular  polygon 
for  the  projection  of  the  normal  system  on  the  yz  plane  is 
oa",  oh",  oc",  and  od" ,  and  Rn'  is  the  projection  of  Rn  on 
that  plane.  Hence  the  action  line  of  Rn  pierces  the  plane 
at  P. 

57.  Principle  of  Moments. — The  moment  of  the  resultant  * 
of  any  system  of  forces  about  a  line  equals  the  sum  of  the 
moments  of  the  forces. 

This  follows  from  the  proposition  that  the  sums  of  the 
moments  of  the  forces  of  equivalent  systems  about  any  line 
are  the  same,  which  we  now  prove. 

Let  /  denote  the  line  or  axis  of  moments  and  imagine  any 
plane  containing  it  as  the  plane  ;:.  Each 
of  the  systems  may  be  compounded  into  a 
resultant  consisting  of  two  forces,  one  in 
and  one  normal  to  the  plane  n;  since  the 
resultants  are  identical,!  Rt  and  Rn  may 
denote  the  forces  of  each  resultant. 

Let  F  (fig.  36)   be  one  of  the  forces  of 
either  system.      The  x  axis  is  taken  along 


Fig.  36. 


the  line  /  and  the  y  axis  in  the  plane  n',  then  the  xy  and  n 


*  Here  "moment  of  the  resultant"  means  the  algebraic  sum  of  the 
moments  of  the  forces  of  the  resultant. 

t  Proof  :  Suppose  that  the}'-  are  not  identical,  and  call  the  forces  of 
the  resultant  of  one  system  Rt'  and  Rn,  and  those  of  the  other  Rt"  and 
Rn".  Being  resultants  of  equivalent  systems,  they  are  equivalent  to 
each  other;  hence  if  the  forces  of  one  resultant  be  reversed,  the  four 
forces  would  balance,  i.e.,  their  resultant  would  be  zero.  Now  the  re- 
sultant of  -R^'and  i?^'' reversed,  is  in  the  plane  n,  so  call  it  Rt"'\  that 
of  Rn  and  Rn"  reversed  is  normal  to  that  plane,  so  call  it  Rn".  But 
the  resultant  of  Rt"'  and  Rn'"  cannot  be  zero  unless  each  is  zero,  and 
that  is  impossible  unless  Rt'  and  Rt"  and  Rn  and  Rn"  are  identical. 


I 


NON-COPLANAR,  NON-CONCURRENT,  NON-PARALLEL  FORCES.     45 


planes  coincide.     F  is  shown  resolved  at  A  into  two  compo- 
nents {Ft  and  Fn)  in  and  normal  to  the  plane  tt.     According 
.  to  art.  28, 

the  mo^iient  of  F  about  /  =  the  moment  of  Fn  about  /; 
hence 

2'(mom.  F)  =  2{Tnom.  Fn). 

But  Rn  is  the  resultant  of  all  the  Fn's,  and,  according  to  art.  53, 

mom.  Rn  about  /  =  2'(mom.  F„)  =^(mom.  F)\ 

hence  the  sum  of  the  moments  of  all  the  forces  of  either  system 
equals  that  of  Rn,  that  is,  the  sums  of  the  moments  of  the 
forces  of  the  equivalent  systems  are  equal.  q.e.d. 

58.  Algebraic  Composition.  —  Determination  of  Rt  and  Rn. 
Fig.  37  represents  one  force,  F,  of  a  system  and  its  components 


Fig.  37. 

Fx,  Fy,  and  Fg,  also  its  components  Ft  and  Fn,  the  plane  ;r 
being  taken  coincident  with  that  of  the  x  and  y  axes.  From  the 
figure,  it  is  plain  that  Fn  equals  F^  and  that  the  x-  and  ;v-com- 
ponents  of  Ft  are  the  same  in  magnitude  and  direction  as  those 
of  F.     It  follows  that 


(i) 


R,  =  JF,     and     R,  =  (2'F/  +  2'F/)^; 


(2)  the  sense  of  Rn  is  given  by  the  sign  of  ^F^; 

(3)  the  angles  which  Rt  makes  with  the  x  and  y  axes  are 

cos-^  IFx/Rt     and     cos-^  IFy/Ru 


46 


EQUiyALENCE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


The  action  lines  of  Rn  and  Rt  are  determined  by  the  principle 
of  moments,  thus:  * 

the  arm  of  Rn  with  respect  to  any  line  in  the  plane  rt  equals 
(the  moment  sum  of  the  given  forces  with  respect  to  that 
line) /Rn; 

the  arm  of  Rt  with  respect  to  any  line  normal  to  tz  equals 
(the  moment  sum  of  the  given  forces  with  respect  to  that 
line)/i^,. 

Determination  of  R  and  C.  Let  F^,  F^,  F^,  etc.,  denote  the 
forces  to  be  compounded  and  O  (fig.  38a)  the  point  through 
which  R  is  to  pass.  The  components  of  Fj  are  the  force  F/  at 
O  and  the  couple  (F^,  i^/')J  the  components  of  F^  are  the  force 
F2'  at  0  and  the  couple  (Fj,  Fg");  ^^c. 


\y 


/: 


/z 


/^ 


M 


-a? 


(a) 


(b) 

Fig.  38. 


18  lb. 


71 


^;/ 1 10  I  J.  J  ^ 


15  1b. 


(c) 


It  is  plain  that  the  components  of  F/,  F^ ,  etc.,  along  any  ■ 
line  are  the  same  in  magnitude  and  direction  as  those  oiF^.F^,^  , 
etc.,  respectively;  hence  \ 


R  =  (2'F,  +2'F^  +^F,y, 
cos  e^  =  i'F^/R,     cos  (^2  =  2'F^/R,     cos  6^  =  i'F^^/R, 

IF x^  IF y,  and  IF^  denoting  the  algebraic  sums  of  the  %,-  y-, 
and  2-components  of  the  given  forces,  and  6^,  d^,  and  6^  the  direc- 
tion angles  of  i?  (art.  50). 

Imagine  the  couple  C  resolved  into  three  components  whose 
planes  are  respectively  perpendicular  to  the  x,  y,  and  z  axes 
(art.  61)  and  denote  them  by  Cx,  Cy,  and  Cg.  Now  the  given 
system  and  the  system  R,  Cx,  Cy,  and  C^  (fig.  386),  have  the 


§IX.]  THEORY  OF  COUPLES.  47 

same  resultant, -and  hence  their  moment  sums  with  respect  to 
any  axis  are  equal.  But  the  moment  sums  of  R,  Cx,  Cy,  and  C^, 
with  respect  to  the  x,  y,  and  z  axes  are  Cx,  Cy,  and  C^.  Hence, 
if  IMx,  ^My,  and  IMz  denote  the  moment  sums  of  the  given 
system  with  respect  to  the  coordinate  axes, 

Cx  =  IMx,     Cy  =  IMy,     and     Cz  =  IM,. 
Also,  by  art.  60,     C  =  (IWx\ iWj  +  IW^Y     and 

cos  ^1  =  JMx/C,     cos  ^2  =  JMj/ZC,     cos  ^3  =  i'M2/C, 
^1,  (f)2,  and  ^3  denoting  the  direction  angles  of  the  vector  of  C. 

EXAMPLES. 

1.  Compound  the  four  forces  of  fig.  38  (c)  into  a  force  at  O 
and  a  couple,  the  edges  of  the  cube  being  4  ft.  long. 

2.  Compound  the  four  forces  into  two  whose  action  lines  are 
i  in  and  normal  to  the  xy  plane. 

§  IX.     Theory  of  Couples. 

59.  Equivalent  Couples. — Proposition. — Two  couples  whose 
imoments,  aspects,  and  senses  are  the  same  are  equivalent;  or, 
otherwise  stated,  two  couples  whose  vectors  are  the  same  are 
equivalent. 

Proof:  I.  The  planes  of  the  couples  coincide.  Let  (Pp)* 
and  (Qq)  (fig.  39)  be  the  two  couples,  F^  (identical  with  Qj), 
F2,  F^  .  .  .  a  system  of  which  (Pp)  is 
the  resultant.  Then  the  vector  sum 
of  Fi,  F^jFs,  .  .  .  F„is  zero,  and  their 
moment  sum  about  any  point  equals 

It  will  now  be  shown  that  the 
couple  (Qq)  is  also  the  resultant  of 
:the  system  F^,  F^j  F^,  ...  by  show-  ^^^-  39- 

ing  that  the  system  can  be  compounded  into  two  forces  which 
are  identical  with  the  forces  of  the  couple.  One  of  those  two 
forces  is  Fj,  and  the  other  one  is  the  resultant  of  F,,  F3,  .  .  . 
which  call  R.     Since  the  vector  sum  of  all  the  forces  of  the  sys- 


*  By  "  couple  (Pp)  "  is  meant  one  whose  forces  are  P  and  arm  is  p. 
For  convenience  of  designation,  the  forces  of  the  couple  are  sometimes 
■marked  P,  and  P,;   then  P,  equals  P». 


48 


EQUIVALET^CE  OF  FORCE  SYSTEMS. 


[Chap.  II. 


tern  is  zero,  R  is  equal  and  opposite  to  F^,  i.e.,  R  is  the  same  in 
magnitude  and  direction  as  Q^.  If  a  denotes  the  arm  of  R  with 
respect  to  an  origin  on  Q^,  then 

i^a  =  moment  sum  of  F2,  F3,   .   .   .    (for  that  origin)  =P^  =  Q^. 

Since  R  equals  Q,  a  =  q,  i.e.,  R  and  Q^  coincide. 

Finally,  since  (Pp)  and  (Qq)  are  the  resultants  of  the  same 
system,  they  are  equivalent. 

II.  The  planes  of  the  two  couples  are  parallel.  Let  (Pp) 
and  (Qq)  (fig.  40)   be  the  two  couples.     According  to  case    I, 

the  couple  (Qq)  can  be  replaced  by 
any  other  in  its  own  plane  pro- 
vided that  its  moment  and  sense 
are  the  same  as  those  of  (Qq).  Let 
(Ss)  be  such  a  couple,  its  forces 
being  parallel  to  those  of  (Pp) 
a>  and  one  in  the  y  axis.  Imagine  a 
system  of  parallel  forces  F^  (iden- 
tical with  Si),  F2,  Fs,  .  .  .  the  re- 
sultant of  which  is  the  couple  (Pp). 
The  vector  sum  of  the  system  is 
zero,  and  its  moment  sums  with 
respect  to  the  x  and  z  axes  equal  the  moments  of  its  resultant 
(Pp)  with  respect  to  the  same  axes,  which  obviously  are  zero 
and  Pp  respectively. 

It  will  now  be  shown  that  the  couple  (Ss),  and  therefore 
(Qq),  is  also  the  resultant  of  F^,  Fj,  Fg,  .  .  .  by  showing  that  the 
system  can  be  compounded  into  two  forces  which  are  identical 
with  the  forces  of  (Ss).  One  of  these  two  forces  is  Fj,  and  the 
other  is  the  resultant  of  F^,  Fg,  ...  which  call  R.  Since  the 
vector  sum  of  all  the  forces  of  the  system  is  zero,  R  is  equal 
and  opposite  to  Fj,  i.e.,  it  is  the  same  in  magnitude  and  direc- 
tion as  Sj.  Let  a  and  c  denote  the  arms  of  R  v/ith  respect  to 
the  X  and  z  axes  respectively ;  then 


i?a  =  moment  sum  of  F^,  Fg,  .   .  .  about  the  x  axis  =  o  | 

and  \ 

Rc=       "  "      "F2,  Fg.  ...       "        '♦    2     "     =Pp=-SS'  \ 

Hence  a  =  o     and     c==s\    i.e.,  R  and  52  coincide. 


§IX.] 


THEORY  OF  COUPLES, 


49 


Finally,  since  {Pp)  and  {Qq)  are  the  resultants  of  the  same 
system,  they  are  equivalent. 

60.  Composition  of  Couples. — Proposition  I.  The  resultant 
of  any  number  of  couples  is  a  couple. 

Proof:  Let  AA'B'B  and  AA"B"B  (fig.  41)  be  the  planes  of 


two  of  the  couples.  Replace  the  couples  by  equivalent  ones 
(F/i)  and  (F/2)  and  so  that  a  force  of  one  couple  shall  "balance" 
a  force  of  the  other;  these  balancing  forces  must  lie  in  AB. 
The  four  forces  are  equivalent  to  two,  F  in  A'B'  and  F  in  A"B", 
which  clearly  constitute  a  couple,  the  resultant  of  the  two  given 
couples.  Similarly,  the  resultant  of  this  resultant  couple  and 
the  third  couple  is  a  couple,  etc. 

The  student  should  supply  a  proof  for  the  case  in  which  the 
planes  of  the  couples  are  parallel. 

Proposition  II.  The  vector  of  the  resultant  of  any  number 
of  couples  is  the  sum  of  the  vectors  of  those  couples. 

Proof:  Consider  first  two  couples,  those  compounded  above. 
Let  CM  and  CN  (fig.  41)  be  their  vectors,  then  CO  is  the  sum 
of  CM  and  CN.  (In  fig.  416,  the  planes  of  the  given  couples 
are  represented  by  their  traces  CC  and  CC"  with  a  plane  per- 
pendicular to  AB.)  It  must  be  shown  that  CO  is  the  vector  of 
the  resultant  couple  {Ff). 

By  construction,  CM /F']^  =  CN /Fj^,  and  the  angles  CNO 
and  C'CC"  are  equal;  hence  the  triangles  CON  and  CC'C"  are 
similar,  and 


CO/j=^CM/j,^CN/U,     or     CO/Ff=^CM/Ff,  =  CN/Ff^; 


50  EQUIVALENCE  OF  FORCE  SYSTEMS.  [Chap.  II. 

that  is,  the  length  of  the  vector  CO  represents  the  moment  of 
the  resultant  couple  to  the  same  ^cale  according  to  which  CM 
and  CN  represent  the  moments  of  the  given  couples.  Since 
CM  is  perpendicular  to  CC ,  CO  is  perpendicular  to  C'C"\  i.e., 
the  vector  CO  is  normal  to  the  plane  of  the  resultant  couple. 
From  an  inspection  of  the  figures,  it  is  plain  that  the  arrow  on 
the  vector  OC  and  the  sense  of  the  resultant  couple  agree  in 
accordance  with  the  rule  of  vector  representation  of  couples 
(see  art.  30).  The  proof  is  easily  extended  to  more  than  two 
couples. 

Special  Cases. — (a)  Three  couples  whose  planes  are  mutually 
at  right  angles.  Let  the  three  planes  be  taken  as  coordinate 
planes  and  call  the  couples  whose  planes  are  perpendicular  to 
the  X,  y^  and  z  axis  Cx,  Cy,  and  Cz  respectively,  C  their  re- 
sultant, and  Vx,  Vy,  Vg,  and  v  their  vectors.     Then 

^  =  (^x^  +  ^2/^  +  ^2^)^J     hence 

Also,  if  ^1,  (j)2,  and  (ji^  denote  the  direction  angles  of  v, 

Qos  ^^  =  Vx/v,     cos  (f)2  =  Vy/v,     cos  (j)3  =  Vg/v; 
hence 

cos^i  =  C^/C,     cos  (l)2  =  Cy/Cy     cos  03  =  €2/0. 

(b)  Couples  whose  aspects  are  the  same.  The  resultant  is 
a  couple  whose  aspect  is  the  same  as  that  of  the  given  couples 
and  whose  moment  equals  the  algebraic  sum  of  the  moments  of 
the  couples,  a  result  reached  in  art.  47  for  coplanar  couples. 

61.  Resolution  of  a  Couple. — It  follows  from  the  preceding 
article  that  a  couple  may  be  equivalent  to  two  or  more  couples, 
which  are  therefore  components  of  that  couple.  Also,  to  re- 
solve a  couple  we  have  only  to  resolve  its  vector,  the  component 
vectors  being  the  vectors  of  the  component  couples. 

The  resolution  of  a  couple  into  three  components  whose 
planes  are  mutually  at  right  angles  is  an  important  special 
case.  Let  C  be  the  couple  to  be  resolved  and  v  its  vector;  and 
denote  the  direction   angles  of  the  vector  by  a,  /?,  and  y,  the 


§IX.]  THEORY  OF  COUPLES.  5^ 

coordinate  planes  having  been  taken  to  coincide  with  the 
planes  of  the  desired  component  couples.  Let  C^,  Cy,  and  Cz 
denote  the  component  couples  which  are  perpendicular  to  the 
X,  y,  and  z  .axes  respectively,  and  v^,  Vy,  and  Vg  the  correspond- 
ing vectors.     Then 

Vx  =  'v  cos  a,     Vy  =  v  cos  1^,     Vg=v  cosyi 
hence 

Cx  =  C  cos  a,     Cy  =  C  cos  /?,     C^  =  C  cos  j. 


EXAMPLES. 

1.  Hold  this  book,  opened  150°,  before  you  and  imagine  two 
couples  whose  moments  are  50  and  70  ft  .-lbs.  to  act  in  the  planes 
of  the  right-  and  left-hand  covers  respectively,  their  senses  being 
clockwise  as  viewed  by  yourself.  On  the  supposition  that  the 
book  is  a  rigid  body,  determine  the  resultant  of  the  two  couples. 

2.  Imagine  two  couples  whose  moments  are  40  and  90  ft.-lbs. 
to  act  in  the  front  and  right-hand  side  of  the  parallelopiped  of 
fig.  15,  their  senses  being  clockwise  and  counter-clockwise  re- 
spectively as  viewed  by  yourself.  Determine  the  resultant  of 
the  four  couples. 


CHAPTER  III.  5 

CENTRE  OF  GRAVITY  AND  CENTROID.  ] 

§  I.     Centroid  of  Parallel  Forces.  ,  1 

62.  Centroid   Defined. — A  system  of  parallel  forces  having  \ 

fixed  application  points  possesses  an  important  property  which  I 

is  now  to  be  investigated.  ! 

First  consider  a  system  of  two  forces,  P  and  Q,  with  appli-  ] 

cation  points  at  A  and  B  respectively,  fig.  25  (a)  or  (b).     In  \ 

art.  43    it  is  shown  that  BC/AC  =  P/Q;  hence  C,  determined  i 

by  the  equation,  is  independent  of  the  angle  between  AB  and  | 

the  action  lines  of  P  and  Q.     Therefore,  if  the  body  upon  which  \ 

the  forces  act  be  turned  in  any  manner,  the  direction  of  the  j 

forces  remaining  unchanged,  their  resultant  will  always  pass  I 

through  the  same  point  of  the  body.     The  point  C  may  hence  \ 

be  regarded  as  the  application  point  of  the  resultant  for  all  | 

aspects  of  the  body.  j 

Consider  next  a  system  of  more  than  two  forces.  j 

Let  F',  F'\  F"\  etc.,  denote  the  forces;  \ 

R'  the  resultant  of  F'  and  F" ,  and  C  its  application  point;  1 

R"  the  resultant  of  R'  and  F'",  and  C"  its  application  ^ 

point;  etc.  i 

If  the  body  be  turned,  R-  always  passes  through  C\  and  R" ,  : 

which  is  also  the  resultant  of  F' ,  F" ,  and  F'",  always  passes  \ 

through  C" .     The  point  C"  may  therefore  be  regarded  as  the  i 

application  point  of   the  resultant  of  F' ,  F" ,  and  F'"  for  all  \ 

aspects  of  the  body.     Extending  this  reasoning  to  the  resultant  \ 

of  the  first  four  force  ■  of  the  system,  then  to  the  resultant  of  j 

the  first  five,  etc.,  one  arrives  at  the  conclusion  that 

the  resultant  of  any  system  of  parallel  forces  having  defi-  I 

52  ^ 


§!•] 


CENTROID  OF  PARALLEL  FORCES. 


53 


nite   application  points   always    passes    through   the    same 
point  of  the  body,  or  its  extension,  irrespective  of  its  aspect. 
This  point  is  called  the  centroid  of  the  system  of  forces. 

63.  Determination  of  the  Centroid. — Call  the  forces  F' ,  F" , 
F'" ,  etc.,  those  of  one  sense  being  given  the  same  sign  and  those 
of  the  opposite  sense  the  opposite  sign.  Let  {x' ,  y' ,z'),  {%" ,  y" , 
z"),  etc.,  denote  the  coordinates  of  their  respective  application 
points  with  respect  to  a  set  of  rectangular  axes  which  is  fixed 
with  reference  to  the  body,  and  let  'x,  y,  z  denote  the  coordinates 
of  the  centroid.  Now  imagine  the  body  upon  which  the  forces 
act  to  be  turned  so  that  one  of  the  axes,  say  that  of  x,  becomes 


F" 

A"               -X" 

y 

J\ 

/ 

* 
^ 

/ 

I      //!                          / 

/ 

^//           \-X"'      '//x- 

"~A 

.A'"  .; 

■7*-. 

^ 

S)a 

/ 

\ 

Fig. 

42. 

parallel  to  the    forces   (see  fig.   42).     Then  according  to  the 
principle  of  moments  (art.  53),  R  being  equal  to  IF , 


^and 


Rz  =  F'z'^F"z"^F"'z'"^  .  . 
Ry^F'y' ^F"y" -^F'"y"' -V  .  . 


Now  imagine  the  body  turned  until  one  of  the  other  axes,  that 
of  y  say,  becomes  parallel  to  the  forces.  The  moment  equation, 
with  the  z  as  moment  axis,  is 

Rx^F'x'^F"x"^F"'x"'-\-  .... 

rFrom  these  three  equations  the  desired  formulas  for  the  coor- 
dinates of  the  centroid  follow;  thus, 

x  =  i'Fx/R,     y=i'Fy/R,     z  =  2'Fz/R. 


54  CENTRE  OF  GRAyiTY  AND  CENTROID.  [Chap.  III. 

In  these  formulas  signs  must  be  given  the  forces  as  explained 
above,  and  to  the  coordinates  of* their  application  points  as 
customarily.  That  this  is  necessary  will  be  seen  from  an  inspec- 
tion of  the  moment  equations  above  and  the  figure. 

//  the  application  points^  of  the  forces  he  coplanar,  two  of  the 
formulas  will  suffice,  provided  that  two  of  the  coordinate  axes 
be  taken  in  the  plane  of  the  application  points.  The  graphic 
method  is  easily  applied  in  this  case  as  follows:  Imagine  the 
body  to  be  turned  so  that  the  action  lines  of  the  forces  fall  into 
the  plane  of  the  application  points,  their  directions  remaining 
unchanged.  The  force  system  is  then  coplanar,  and  the  action 
line  of  its  resultant  may  be  readily  determined  graphically 
(art.  39).  Now  imagine  the  body  to  be  turned  about  an  axis 
which  is  perpendicular  to  the  plane  of  the  application  points 
through  any  angle  other  than  180  or  360  degrees.  The  forces 
are  still  coplanar,  and  the  action  line  of  their  resultant  may  be 
determined  as  before.  The  intersection  of  the  action  lines  of 
the  two  resultants  is  the  centroid  of  the  system. 

EXAMPLES. 

1.  Forces  of  10,  20,  15,  and  5  lbs.  have  the  same  directions; 
the  application  points  are  coplanar  and  their  coordinates  in  feet 
are  respectively,  (5,  3),  (2,  4),  (i,  5),  and  (5,  6).  Determine  the 
centroid  of  the  forces.  Ans.  ^  =  2.6  ft. 

2.  With  the  forces  of  the  preceding  example,  include  two  of 
20  and  30  lbs.  whose  directions  are  the  same  as  that  of  the  four 
and  whose  application  points  are  respectively  at  (5,  4)  and 
(  —  6,  3).     Determine  the  centroid  of  the  six  forces. 

Ans.  ^  =  0.5  ft. 

3.  Reverse  the  sense  of  the  thirty-pound  force,  and  deter- 
mine the  centroid  of  the  six. 

§  II.     Centre  of  Gravity  of  a  Body. 

64.  Definition  and  General  Formulas. — The  weights  of  the 
particles  of  a  body  constitute  a  system  of  practically  parallel 
forces  having  fixed  application  points.  Therefore,  these  forces 
have  a  centroid,  that  is,  the  resultant  always  passes  through  a 
certain  point  fixed  with  reference  to  the  body  no  matter  how  it 


■    §11.]  CENTRE  OF  GRAVITY  OF  A  BODY.  55 

is  turned.  It  is  assumed,  of  course,  that  in  turning  the  body, 
its  form  and  size  are  not  changed. 

Definition. — The  centre  of  gravity  of  a  body  is  the  centroid 
of  the  weights  of  all  its  particles. 

Centres  of  gravity,  in  the  following,  will  be  usually  specified 
by  means  of  rectangular  coordinates  and  which  will  then  always 
be  denoted  by  x,  y,  and  'z.  General  values  for  these  may  be 
deduced  as  follows:  Let  {x\  y' ,  z'),  {%" ,y" ,  2"),  etc.,  denote  the 
coordinates  of  the  particles*  of  a  body;  w' ,  w" ,  etc.,  their 
weights,  and  W  the  weight  of  the  body.  Then,  from  the 
formulas  for  centroid  of  a  system  of  parallel  forces, 

x  =  2'wx/W,     y  =  i'wy/W,     z  =  i'wz/W. 

65.  Moment  of  a  Weight  with  Respect  to  a  Plane. — De-fini- 
iion. — The  moment  of  the  weight  of  a  body  with  respect  to  a 
plane  is  the  product  of  the  weight  and  the  ordinate  of  the  centre 
of  gravity  of  the  body  from  that  plane.  Ordinates  on  opposite 
sides  of  the  plane  are  given  opposite  signs.  A  moment,  there- 
fore, has  the  same  sign  as  the  corresponding  ordinate. 

From  the  definition  it  follows  that  if  the  moment  of  the 
weight  of  the  body  is  zero  with  respect  to  a  plane,  its  centre  of 
gravity  is  in  that  plane. 

Proposition. — The  moment  of  the  weight  of  a  body  with 
respect  to  a  plane  equals  the  algebraic  sum  of  the  moments  of 
the  weights  of  its  parts  with  respect  to  the  same  plane. 

Proof:  Let  W^^,  W2,  etc.,  be  the  weights  of  the  parts  and  x^. 
x^\  etc.,  be  the  ic-coordinates  of  particles  of  the  first  part,  and 
«;/,  w/',  etc.,  their  weights;  x^ ,  x^\  etc.,  be  coordinates  of 
particles  of  the  second  part,  and  w^ ,  w^' ,  etc.,  their  weights,  etc. 
Then,  according  to  the  formulas  above,  the  :x:-coordinate  of  the 
centre  of  gravity  is 

__(WiV4-w/^^/^+  .  .  .)  +  (w2V+7£;/V^+  .  .  .)  +  etc. 
(w/+«^/'+   .  .  .)  +  (w/+2£;2"+   •  •  O+etc. 

*  By  particle  is  meant  a  body  whose  dimensions  are  vanishingly  small 
or  negligible  in  comparison  with  other  distances  involved,  which,  in  this 
case,  are  for  each  particle  the  distances  between  it  and  the  coordinate 
planes. 


56 


CENTRE  OF  GRAyiTY  AND  CENTROID.  [Chap.  III. 


Now  the  first  parenthesis  of  the  numerator  equals  W^^,  the 
second  parenthesis  equals  W^^,  elc.,  and  the  denominator 
equals  the  weight  of  the  entire  body.  The  equation  may  there- 
fore be  written 

Wx=WiXi-fW2X2+   .  .  . 

Since  the  plane  from  which  the  x's  are  measured  may  be  taken 
at  pleasure,  the  proposition  is  proved. 

EXAMPLES. 

I.  Suppose  that  the  heavy  lines  of  fig.  43  represent  a  bent 
wire.     Determine  the  coordinates  of  its  centre  of  gravity. 

Ans.  3^  =  4.13  in. 

\y 

bI 


Fig.  43- 


Fig.  44. 


2.  A  piece  of  tin  consists  of  three  parts,  square,  semicircular, 
and  equilateral  triangular  (fig.  44).  Determine  the  centre  of 
gravity  of  the  piece.  (The  distances  of  the  centres  of  gravity 
of  the  semicircular  and  triangular  parts  from  the  base  are  4/3;r 
times  the  radius  and  -J  times  the  altitude  respectively.) 

3.  Imagine  the  triangular  and  circular  parts  described  in  the 
preceding  example  bent  forward  on  the  lines  BC  and  AC  re- 
spectively until  their  planes  are  perpendicular  to  that  of  the 
square.     Determine  the  centre  of  gravity. 

4.  Suppose  that  the  triangular  part  of  the  preceding  example 
is  bent  backward  instead  of  forward.  Write  the  expressions 
for  the  coordinates  of  the  centre  of  gravity. 

5.  The  weights  of  four  bodies  sxe  W^,  W^,  M'V  and  W^,  and 
the  distance  of  the  centre  of  gravity  of  the  first  from  the  plane 
through  the  centres  of  gravity  of  the  other  three  is  h.  How  far 
from  that  plane  is  the  centre  of  gravity  of  the  four  bodies? 

6.  The  weights  of  three  bodies  are  12,  18,  and  40  lbs.;  the 


§11.] 


CENTRE  OF  GRAyiTY  OF  A  BODY, 


57 


distances  between  the  centres  of  gravity  of  the  first  and  second, 
second  and  third,  third  and  first,  are  lo,  14,  and  16  inches  re- 
spectively. How  far  from  the  line  joining  the  centres  of  gravity 
of  the  first  and  second  is  the  centre  of  gravity  of  the  three  ? 

Ans.  7.91  in. 

7.  Fig.  45  represents  a  cylindrical  body  having  a  cylindrical 
hole  in  the  top ;  a  part  of  the  body  is  cast  iron 
and  the  remainder  (conical)  is  lead.     Determine 
the  centroid.      (Cast  iron  and  lead  weigh  450 
and  71 1  lbs.  per  cu.  ft.  respectively.    See  art.  80.) 

66.  Centre  of  Gravity  Determined  by  Integra- 
tion. —  Imagine  a  body  divided  into  an  infi- 
nitely large  number  of  parts,  i.e.,  elements. 
Let  dW  denote  the  weight  of  any  element  and 
X,  y,  z,  the  coordinates  of  its  centroid.  Accord- 
ing to  the  principle  of  moments, 

Wx=JdW-x,  Wy=ydW.y,Wz  =  y*dW.z.  (i) 

If  the  body  is  homogeneous,  let  w  denote  its 
specific  weight  *  and  V  its  volume ;  then  W=wV  and  dW  =wdV. 
Equations  (i)  reduce  to 


Vx  =  y*dV.x,     Vy  =  ydV.y,     Vz==/dV.z. 


(2) 


These  formulas  may  be  employed  for  determining  the  centre  of 
gravity  of  a  body  which  cannot  be  divided  into  finite  parts  whose 
weights  and  centres  of  gravity  are  known,  provided  that  its 
form  and  specific  weight,  if  the  body  is  not  homogeneous,  are 
such  that  the  integrations  can  be  performed. 

EXAMPLES. 

I.  Determine  the  centre  of  gravity  of  an  octant  of  a  sphere 
whose  specific  weight,  varying  from  point  to  point,  is  directly 
proportioned  to  distance  from  the  centre. 

Solution:  Let  th^  plane  faces  of  the  octant  be  taken  as 
coordinate  planes  as  shown  in  fig.  46 ;  denote  the  radius  of  the 


*  By  specific  weight  is  meant  weight  per  unit  volume. 


58 


CENTRE  OF  GR/fJ^ITY  AND  CENTROID.  [Chap.  III. 


Sphere  by  r  and  the  specific  weight  at  any  point  P  by  w.     If  k 
is  a  proper  constant  and  x,  y,  and  z  denote  the  coordinates  of  P^ 

w  =  k{x^+y^+z^)^     and     dW  =  k{x^+y^+z^ji  dx  dy  dz. 
Then  from  (i), 

Wx=k/      /  J^        (x^+y^+z^)^dxdydz'X; 

hence  x  =  ^r.     E vddently ,  x=y=z. 


Fig.  46.  ; 

2.  Determine  the  centre  of  gravity  of  an  octant  of  a  homo-  ; 
geneous  sphere. 

Solutions:  Equation  (2)  maybe  employed.     First,  selecting 

a  cubical  element  as  in  ex.  i,  dV  =  dx  dy  dz,  and  eq.  (2)  becomes  ; 

Vx=  II               I        dx  dy  dz'X\  ■« 

hence         -             x=^r.     Evidently,  x=y=l.  ;j 

Second,    selecting    the    parallelopiped    AB    as    elementary  ' 

volume,  dV  =  dx  dy{AB).     If  x  and  y  denote  coordinates  of  By  | 

AB  =  {r^-x^-y^)^,  | 

and  eq.  (2)  becomes  J 


hence 


Vx=/    7         ir^-x^''y^)^dxdy'X\ 


x=ir. 


§11.]  CENTRE  OF  GRAVITY  OF  A  BODY,  59 

Third,  selecting  the  volume  between  two  planes  parallel  to 
the  yz  plane,  dx  apart  as  elementary  volume,  dV  ^izDC  dxj \. 
If  X  denotes  the  coordinate  of  C,  DC  ^r'^  —  x'^,  and  eq.  (2)  be- 
comes 

Vx=—  I    (r^  —  x^)dx'X] 
hence  x=ir.  . 

67.  Centre  of  Gravity  Determined  Experimentally. — Some 
bodies  are  so  irregular  in  shape  that  their  centres  of  gravity  can- 
not be  found  by  the  methods  explained  above.  In  such  cases, 
experimental  methods  may  be  resorted  to. 

The  Method  of  Suspension. — The  body  whose  centre  of 
gravity  is  to  be  determined  is  suspended  from  one  point  of  it  and 
the  direction  of  the  suspending  cord  is  then  marked  in  some  way 
on  the  body.  The  operation  is  repeated  using  another  point  of 
suspension.  Since  the  centre  of  gravity  is  in  each  of  the  lines 
so  fixed  in  the  body,  it  is  at  their  intersection. 

The  Method  of  Balancing. — The  body  whose  centre  of  gravity 
is  to  be  determined  is  balanced  upon  a  straight-edge  and  the 
position  of  the  vertical  plane  containing  the  edge  marked  upon 
the  body ;  then  the  operation  is  repeated  for  two  more  balancing 
positions  of  the  body.  Since  the  centre  of  gravity  is  in  each 
plane  so  fixed,  it  is  in  the  common  point  of  the  three. 

This  method  is  easily  applied  to  determine  the  centre  of 
gravity  of  a  thin  plate.  The  plate  is  balanced  in  two  positions, 
same  face  down  both  times,  and  the  lines  of  contact  of  the 
straight-edge  and  plate  are  marked  upon  the  latter.  The  centre 
of  gravity  is  midway  between  the  intersection  of  those  lines  and 
a  point  directly  opposite  on  the  other  face  of  the  plate. 

EXERCISE. 

Cut  from  a  sheet  of  stiff  paper  the  ' ' angle  section"  described 
in  ex.  4,  art.  81,  and  determine  how  far  the  centre  of  gravity  of 
the  paper  is  from  the  edges  AC  and  CB. 


6o  CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III. 

§  III.     Centroids  of  Solids,  ^Surfaces,  and  Lines. 

68.  Centroid  Defined. — The  term  centre  of  gravity,  applied 
to  solids,*  surfaces,  or  lines,  is  inappropriate,  for  they  have  no 
weight.  However,  the  term  is  much  used  in  that  connection. 
Centroid,  a  more  suitable  term,  is  coming  into  use.  Instead  of 
the  terms  centroid  or  centre  of  gravity  of  a  solid  surface  or  line, 
centroid  or  centre  of  gravity  of  a  volume  area  or  length,  re- 
spectively, are  often  employed. 

Definitions. — The  centroid  of  a  solid  is  that  point  of  it  which 
coincides  with  the  centre  of  gravity  of  a  homogeneous  body  which 
is  bounded  by  the  surface  of  the  solid.  The  centroid  of  a  surface 
is  the  limiting  position  of  the  centre  of  gravity  of  a  homogeneous 
thin  plate  one  of  whose  faces  coincides  with  the  surface  as  its 
thickness  approaches  zero.  The  centroid  of  a  line  is  the  limiting 
position  of  the  centre  of  gravity  of  a  homogeneous  slender  rod 
whose  axis  coincides  with  the  line  as  its  sectional  area  approaches 
zero. 

69.  Centroid  as  Mean  Point. — Proposition. — The  ordinate 
of  the  centroid  of  any  solid,  surface,  or  line  with  reference  to  any 
plane  equals  the  mean  ot  the  ordinates  of  all  the  equal  elements 
of  the  solid,  surface,  or  line.  It  must  be  understood  that  oppo- 
site signs  are  given  to  ordinates  on  opposite  sides  of  the  plane. 

Proof  for  lines:  Let  x^^  x^,  x^,  etc.,  denote  the  ordinates  to 
the  different  elementary  lengths  ds  of  the  line,  and  nthe  num- 
ber of  elements  (infinite).     Then 

,  rXTj  +  ^2  I  -^3  '  •  •  •  ,       • 

the  mean  ordmate  =  — ^ , 

n 

.    ^    .  .            ^       (X  +x,+x,+  ...)ds     f^^^ 
but  this  equals     ^-^^ ^ —  = — 1 — . 

/  denoting  the  length  of  the  curve.  In  art.  83  it  is  shown  that  the 
last  expression  is  that  for  the  ordinate  of  the  centroid,  hence  the 
proposition  is  proved. 

Proof  for  solids  and  surfaces  is  similar  to  that  just  given. 


*  Geometrical  and  not  physical  solid  is  meant. 


§IIL]  CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES.  6i 

70.  Moment  of  a  Volume,  Area,  or  Length. — Definitions. — 
The  moment  of  the  volume  of  a  solid  (area  of  a  surface  or  length 
of  line)  with  respect  to  a  plane  is  the  product  of  the  volume 
(area  or  length)  and  the  ordinate  of  the  centroid  of  the  solid 
(surface  or  line)  from  that  plane.  Ordinates  on  opposite  sides 
of  the  plane  are  given  opposite  signs;  hence  a  moment  has  the 
same  sign  as  the  corresponding  ordinate. 

Proposition. — The  moment  of  the  volume  of  any  number  of 
solids  with  respect  to  a  plaiie  equals  the  algebraic  sum  of  the 
moments  of  the  volumes  of  those  solids  with  respect  to  the  same 
plane.     Similar  propositions  hold  for  surfaces  and  lines. 

Proof  for  volumes:  Let  V^,  V^,  etc.,  denote  the  volume  of  the 
solids,  Wj,  U2,  etc.,  the  ordinates  of  their  centroids,  and  w  that  of 
the  centroid  of  the  collection  of  solids  with  respect  to  any  plane. 
Now  the  ordinates  of  the  centres  of  gravity  of  homogeneous 
bodies  which  are  bounded  by  the  surfaces  of  these  solids  are  also 
u^,  U2,  etc.,  and  the  ordinate  of  the  centre  of  gravity  of  the 
collection  of  bodies  is  also  u.  From  art.  65,  if  w  denotes  the 
specific  weight  of  the  imagined  homogeneous  bodies, 

{wVi+wV2+  .  .  .)u  =  wV^-u^+wV2'U2-\-  •  •  •  > 
or 

(V1+V2+   .  .  .  )u  =  ViUi+V2U2+   .  .  .  Q.E.D. 

Proof  for  surfaces:  Let  A^,  A^,  etc.,  denote  the  areas  of  the 
surfaces,  u^,U2,  etc.,  the  ordinates  of  their  centroids,  and  w  the 
ordinate  of  the  centroid  of  the  group.  Imagine  now  as  many 
homogeneous  thin  plates  as  there  are  surfaces,  and  that  one  face 
of  each  plate  coincides  with  one  of  the  surfaces.  Let  w/,  U2  , 
etc.,  denote  the  ordinates  of  the  centres  of  gravity  of  the  plates, 
and  u'  that  of  the  centre  of  gravity  of  the  collection.  If  w 
denotes  the  specific  weight  of  the  plates  and  t  their  thickness, 
their  weights  are  approximately  Aj,w,  A2tw,  etc.,  the  approxi- 
mation being  closer  the  smaller  /  is  taken.  (Of  course,  for  plane 
plates  these  expressions  are  correct  for  all  values  of  ^)  From 
art.  65, 

(.4i+^2+  •  •  .)twu'  =A^tw-u^' -\-A2tW'U2'  +  .  .  .,  approximately; 


^2  CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III. 


or 


(Aj^+A2+  .  .  .)u'=A^Mi  +A2U2*r^.  .  .  ,  approximately. 


Now  this  approximate  equality  approaches  exact  equality  as  t 
approaches  zero,  i.e., 

lim.  [(^1+^2+  .  .  .)w']=lim.  [A^il^' +A2U2' +  .  .  .] 

=  lim.  [/liw/]  +  lim.  [.42^2']+  .  .  .  , 
or 

(A1+A2+  .  .  .)u  =  AiUi+A2U2+  .  .  .      ,  Q.E.D. 

Proof  for  lines  is  very  similar  to  that  for  areas. 

71.  Centroidal  Plane. — Definition. — Any  plane  containing 
the  centroid  of  a  line,  surface,  or  solid  will  be- called  a  centroidal 
plane  of  that  line,  surface,  or  solid.* 

From  the  definitions  of  art.  70  it  follows  that  if  the  moment 
of  a  length,  area,  or  volume  with  respect  to  a  plane  is  zero, 
then  it  is  a  centroidal  plane  of  the  line,  surface,  or  solid. 

Proposition. — If  the  form  of  a  solid  (surface  or  line)  is  such 
that  it  can  be  divided  into  parts  which  may  be  paired  off  in  such 
a  way  that  the  parts  of  each  pair  are  equal  in  volume  (area  or 
length),  and  that  the  lines  joining  the  centroids  of  the  parts  of 
each  pair  are  bisected  by  a  plane,  then  that  plane  is  a  centroidal 
one. 

Proof:  The  moments  of  the  volumes  (areas  or  lengths)  of  the 
parts  constituting  a  pair  with  respect  to  the  bisecting  plane  are 
equal  but  of  opposite  sign;  hence  the  moment  of  the  volume 
(area  or  length)  of  the  pair  is  zero,  and  the  moment  of  the  entire 
volume  (area  or  length)  is  zero ;  therefore  that  plane  is  one  of  zero 
moment  audit  contains  the  centroid  of  the  solid  (surface  or  line). 

Evidently  the  following  are  centroidal  planes: 
for  a  circular  arc,  any  bisecting  plane  containing  a  diameter; 
for  a  sector,  any  bisecting  plane  containing  a  diameter; 
for  a  triangle,  any  plane  cutting  it  in  a  median; 
for  a  parallelogram,  any  plane  cutting  it  in  a  diagonal; 
for  a  triangular  pyramid,  any  plane  containing  its  vertex  and 

median  of  the  base. 

*  Likewise  any  straight  line  containing  the  centroid  of  a  line,  surface, 
or  solid  will  be  called  a  centroidal  axis  of  the  line,  surface,  or  solid. 


§  III.]        CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES.  63 

72.  Centroids  of  Simple  Solids  and  Surfaces. — The  centroids 
of  such  solids  and  surfaces  are  determined  in  the  succeed- 
ing articles.  The  general  method  consists  in  finding  enough 
planes  or  lines  containing  the  centroid  to  locate  its  posi- 
tion. 

73.  The  Centroid  of  a  Triangle  is  at  the  intersection  of  its 
medians. 

Proof:  As  before  stated,  a  plane  cutting 
a  triangle  in  a  median  is  a  centroidal  plane. 
Since  the  centroid  is  in  such  a  plane  and  in 
the  plane  of  the  triangle,  it  is  in  their  inter- 
section, i.e.,  the  median  line.  But  there  are 
three  such  median  lines,  hence  the  centroid  is 
at  their  intersection.  Further,  OA' ,  OB',  and 
OC  (fig.  47)  equal  respectively  one-third  of  AA\  BB\  and 
CC ,  from  which  it  follows  that 

the  distance  of  the  centroid  of  a  triangle  from  any  side 
equals  one-third  the  altitude  measured  from  that  side. 

74.  The  Centroid  of  a  Parallelogram  is  at  the  intersection  of 
its  diagonals. 

Proof:  As  before  stated,  a  plane  cutting  the  parallelogram 
in  a  diagonal  is  a  centroidal  plane.  Since  the  centroid  is  in 
such  a  plane  and  in  the  plane  of  the  parallelogram,  it  is  in  their 
intersection,  the  diagonal;  hence,  etc. 

75.  The  Centroid  of  the  Surface  of  a  Pyramid  is  on  the  axis  * 
of  the  surface  at  a  distance  from  the  base  equal  to  one-third  of 
the  altitude. 

Proof :  Imagine  the  pyramid  cut  by  numerous  planes  parallel 
to  its  base;  the  part  of  the  surface  between  any  two  adjacent 
planes  is  a  frustum  of  the  surface.  Conceive  the  entire  surface 
as  consisting  of  elementary  frustums;  the  centroid  of  any  one 
of  them  and  those  of  the  perimeters  of  its  bases  approach 
coincidence.  But  these  perimeters  are  polygons  similar  to 
the  perimeter  of  the  base  and  the  relation  between  any 
one    of   them    and    the    intersection    of    its   plane    with    the 

*  By  axis  of  the  surface  of  a  pyramid  (or  cone)  is  meant  the  line  join- 
ing its  apex  with  the  centroid  of  the  perimeter  of  its  base. 


64  CENTRE   OF  CRAyiTY  AND  CENTROID,         [Chap.IIL 

axis  is  precisely  similar  to  that  between  the  perimeter  of 
the  base  and  its  centroid.  Henoe  the  centroid  of  all  the 
polygons  and  the  elementary  frustums  lie  upon  the  axis,  and 
it  follows  that  the  centroid  of  the  surface  of  the  pyramid  is  on 
the  axis. 

The  centroids  of  all  the  faces  of  the  pyramid  lie  in  a  plane 
distant  one-third  of  the  pyramid's  altitude  from  the  base.  It 
follows  that  the  plane  is  a  centroidal  one,  and  hence  the  centroid 
of  the  surface  is  in  that  plane. 

76.  The  Centroid  of  the  Surface  of  a  Cone  is  on  the  axis  of 
the  surface  at  a  distance  from  the  base  equal  to  one-third  the 
altitude. 

Proof :  The  surface  of  a  cone  may  be  regarded  as  the  limit  of 
the  surface  of  a  pyramid,  the  number  of  whose  faces  are  in- 
creased without  limit.  It  follows  ihat  the  limiting  position  of 
the  centroid  of  the  pyramid  is  the  centroid  of  the  surface  of  the 
cone,  etc. 

77.  The  Centroid  of  a  Prism  with  Parallel  Bases  is  in  the 
axis  and  midway  between  the  bases. 

Proof:  Conceive  the  prism  as  consisting  of  elementary 
laminas  whose  faces  are  parallel  to  the  bases.  The  centroid  of 
any  lamina  and  the  centroids  of  its  faces  approach  coincidence. 
Obviously,  the  centroids  of  the  faces  are  on  the  axis,  hence  the 
centroid  of  each  elementary  lamina  is  also,  and  it  follows  that 
the  centroid  of  the  prism  is  on  the  axis.  Obviously,  a  plane 
midway  between  the  bases  is  a  centroidal  one,  hence  the  centroid 
is  in  that  plane. 

78.  The  Centroid  of  a  Pyramid  with  a  Triangular  Base  is  on 
the  axis  *  at  a  distance  from  the  base  equal  to  one- fourth  the 
altitude  (see  also  next  art.). 

Proof:  As  before  stated,  a  plane  containing  the  apex  and  a 
median  line  of  the  base  is  a  centroidal  one.  But  there  are 
three  such  planes,  and  since  the  centroid  is  in  each,  it  is  in  their 
intersection,  the  axis.  Now  each  '  *  corner  "  of  the  pyramid  may 
be  in  turn  considered  as  a  vertex  corresponding  to  which  there 


*  A  line  joining  the  apex  of  any  pyramid  or  cone  with  the  centroid  of      1 
the  base  is  the  axis.  i 


§  III.]  CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES.  C5 

is  an  axis,  and  the  centroid  being  on  each  axis,  is  at  their  in- 
tersection. 

Two  axes,  AF  and  BG,  are  represented  in 
fig.  48.  Since  EF  =  EB/s  and  EG  =  EA/s, 
GF is  parallel  to  AB  and  GF  =  AB/3,  and  it 
follows  that  the  triangles  OFG  and  OAB  are 
similar.  Hence  OF  =  OA/s  and  0F  =  AF/4, 
that  is,  the  centroid  O  is  one-fourth  the 
length  of  the  axis  upward  from  its  foot.  ^^ 
Also  the  distance  of  the  centroid  from  the 
base  equals  one-fourth  the  altitude. 

79.  The  Centroid  of  Any  Pyramid  is  on  the 
axis  at  a  distance  from  the  base  equal  to  one-fourth  the  altitude. 

Proof :  Conceive  the  entire  pyramid  as  made  up  of  elementary 
frustums;  the  centroid  of  any  one  of  them  and  those  of  its 
faces  approach  coincidence.  But  these  faces  are  surfaces 
similar  to  the  base,  and  the  relation  between  any  one  of  them  and 
its  intersection  with  the  axis  is  precisely  similar  to  the  relation 
between  the  base  and  its  centroid.  Hence  the  centroids  of  all 
such  surfaces  and  elementary  frustums  lie  upon  the  axis,  and 
it  follows  that  the  centroid  of  the  pyramid  is  in  the  axis. 

Conceive  the  base  divided  into  triangles,  and  the  entire 
pyramid  as  consisting  of  pyramids  with  these  triangles  as  bases. 
The  centroids  of  all  these  component  pyramids  lie  in  a  plane 
distant  one-fourth  of  the  pyramid's  altitude  from  the  base;  it 
follows  that  the  plane  is  a  centroidal  one  for  the  entire  pyramid, 
and  hence  the  centroid  is  in  that  plane. 

80.  The  Centroid  0}  a  Cone  is  in  its  axis  at  a  distance  from 
the  base  equal  to  one-fourth  the  altitude. 

Proof:  A  cone  may  be  regarded  as  the  limit  of  a  pyramid, 
the  number  of  whose  faces  are  increased  without  limit.  It 
follows  that  the  limiting  position  of  the  centroid  of  the  pyramid 
is  the  centroid  of  the  cone.     Hence,  etc. 

81.  Centroids  of  Solids  and  Surfaces  Consisting  of  Simple 
Parts. — The  solids  and  surfaces  whose  centroids  are  now  to  be 
determined  consist  of  parts  whose  volumes,  or  areas,  and  cen- 
troids are  known.  The  solutions  are  based  on  the  principle  of 
moments,  art.  70. 


66 


CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III. 


EXAMPLE^ 

I.  Determine  the  centroid  of  a  trapezoid  whose  altitude  is 
a  and  whose  minor  and  major  bases  are  b  and  B  respectively. 

Solution:  Imagine  the  trapezoid  divided  into  two  triangles. 
Their  areas  are  Ba/2  and  ba/2,  and  the  distances  from  the  base 
B  to  their  centroids  are  a/^  and  20/3  respectively.  The  area  of 
the  trapezoid  is  {B  +  b)a/2;  and  if  y  denotes  the  distance  be- 
tween its  centroid  and  the  base  B, 


or 


(B  +  b)a_     Ba  a     ba  2 

^ -y  = + a, 

2       -"       23      23 

_      2b+B 


The  centroid  can  be  determined  geometrically  in  this  way: 
Extend  in  either  direction  the  major  base  a  distance  b,  and  in 
the  opposite  direction  the  minor  base  a  distance  B.  Then  the 
line  joining  the  ends  of  the  extensions  intersects  the  line  joining 
the  centres  of  the  bases  at  the  centroid  of  the  trapezoid.  The 
student  should  supply  a  proof. 

2.  Determine  the  centroid  of  the  "tee  section"  represented 
in  fig.  49.  Ans.  i.oi  in.  above  base. 

3.  Determine   the  centroid  of  the  ** channel  section"  repre 
sented  in  fig,  49.  Ans.  0.79  in.  above  base. 


f0.40 


'Tee' 


^% 


i^ 


31/^ 


-3K— - 


B' 

"Angle" 

c  > 

7i 

A' 

C 

— ^ — 

—J. 

<- 

"Channel" 


7    ^' 


Y      ^0.90 


;,  0  40 


0.90^ 


15inr- 

FlG.   49. 


4.  Determine  the  centroid  of  the  "angle  section"  repre- 
sented in  fig.  49,  with  reference  to  the  sides  AC  and  BC. 

5.  Locate  the  centroid  of  any  plane  quadrilateral. 
Method:  Join  the  centroids  of  the  two  triangles  into  which 

either  diagonal  divides  it;  also  join  the  centroids  of  the  triangles 


§IIL] 


CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES. 


67 


into  which  the  other  diagonal  divides  it.     The  intersection  of 
those  two  lines  is  the  centroid  sought.     Prove. 

82.  Centroids  of  Solids  and  Surfaces  Considered  as  Parts  of 
Other  Solids  or  Surfaces. — The  solids  and  surfaces  whose  cen- 
troids are  now  to  be  determined  may  be  conveniently  con- 
sidered as  consisting  of  a  simple  solid  or  surface  minus  one  or 
more  simple  solids  or  surfaces.  It  is  supposed  that  the  volume 
or  area  and  centroid  of  each  of  the  simple  solids  or  surfaces  are 
known.  The  principle  of  moments  slightly  modified  is  em- 
ployed thus:  Let  M^  be  the  moment  of  the  volume  or  area  in 
question,  M2  that  of  the  volume  or  area  of  which  the  first  is  a 
part,  and  M^  that  of  the  remainder;  then 

for,  according  to  the  principle  of  moments,  M2  =  M^-\-M^. 

EXAMPLES. 

I .  Determine  the  centroid  of  the  shaded  part  of  fig.  50. 

Solution:    The  shaded  area  consists  of    ^__a/^___^_,a/„ ^ 

the   large   square  minus  the  triangle  and 
quadrant.     Its  value  is 

,2 
;(l2-7r), 


Let  the  base  and  left  side  of  the  square 

respectively  be  x  and  y  axes,  and  x  and  y 

the  coordinates  of  the  centroid.     The  co-  Fig.  50. 

ordinates  of  the   centroid   of  the   triangle   and    quadrant   are 

respectively 

(a/6,  2 a/3)     and     (a  — 2a/3;r,  2a/3;r).    (See  ex.  4,  art.  83.) 
Hence 


16 


or 


— (i2-7r)-^=  a}-----~-^{a-2alziz), 


_     8-r 
x= a. 


12 


The  student  should  determine  "y, 

2.  Two  circles  whose  diameters  are  as  2  to  3  are  tangent 
internally.  Locate  the  centroid  of  the  part  of  the  larger  circle 
not  included  in  the  smaller.  ' 


68  CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III. 

3.  Locate  the  centroid  of  a  surface  bounded  by  a  quarter 
of  a  circle  and  the  tangents  at  its  extremities. 

Ans.  Its  distance  from  either  tangent  =o.22^r. 

4.  Suppose  that  in  fig.  49  three  comers  of  the  "angle"  are 
rounded  as  shown,  the  radii  at  /I'  and  B'  being  -f^-  in.  and  that 
at  C  i  in.     Redetermine  the  centroid. 

5.  Locate  the  centroid  of  a  frustum  of  a  cone  whose  major 
and  minor  base  radii  are  R  and  r  respectively  and  whose  altitude 

^  ''■  Ans.  Distance  from  larger  base  is  ''-?^^1^±^. 

6.  Locate  the  centroid  of  the  shaded  part  of  fig.  51. 
Solution:     Since   any    plane    cutting   the 

figure  in  Ox  is  a  centroidal  plane,  the 
centroid  is  on  Ox  and  y  =  o.  To  determine 
X,  consider  the  shaded  part  as  consisting  of 
the  sector  of  radius  r2  minus  the  sector  of 
Fig.  51.  radius   n,   and   write    the   moment   equation 

with  respect  to  the  yz  plane.     (See  ex.  3,  art.  83.) 

Ans.x^^^rlzi^^ir,^, 
3(^2-^i)«        2 

7.  Locate  the  centroid  of  a  circular  segment. 

Ans.  Its  distance  from  the  centre  of  the  circle  equals 

(base  of  the  segment) V(3-rea  of  the  segment)  12. 

83.  Centroids  Determined  by  Integration. — Imagine  the  solid, 

surface,  or  line   divided   into   elementary  parts.     Let  'x,  y,  z 

denote  the  coordinates  of  the  centroid  of  the  solid,  surface,  or 

line. 

For  a  solid,  let  V  denote  its  volume  and  x,  y,  z  the  coordi- 
nates of  the  centroid  of  any  elementar^^'part.  Then,  according 
to  the  principle  of  moments"  (art.  70), 

Vx  =  /dVx,     Vy  =  /dV.y,     Vz  =  /dV.z.    .     .     (i) 

For  a  surface,  let  A  denote  its  area,  and  x,  y,  z  the  coordi- 
nates of  the  centroid  of  any  elementary  part.  According  to 
the  principle  of  moments, 

Ax  =  ydAx,      Ay  =  y*dAy,     Az=y'dAz.    .     .     (2) 


5  III. 


CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES. 


69 


For  a  line,  let  /  denote  the  length,  and  x,  y,  z  the  coordinates 
of  the  centroid  of  any  elementary  part.  According  to  the  prin- 
ciple of  moments, 

[      '.  lx  =  /dlx,    ly  =  /dl.y,    lz  =  /dlz.    ...     (3) 

The  limits  of  integration  in  the  above  formulas,  applied  in 
any  particular  example,  must  be  such  that  the  moments  of  each 
element  of  the  solid  (surface  or  line)  are  included  in  the  summa- 
tions. The  formulas  may  be  used  for  determining  the  centroid 
of  a  solid  (surface  or  line)  provided  that  the  form  of  it  is  such 
that  the  integrations  can  be  performed.  They  are  to  be  em- 
ployed when  the  solid  (surface  or  line)  cannot  be  divided  into 
parts  whose  volumes  (areas  or  lengths)  and  centroids  are  known. 


EXAMPLES. 

I.  Locate  the  centroid  of  a  circular  arc. 


Solution:  Evidently  the  centroid  is  on  OC  (fig.  52),  i.e., 
y  =  o.  To  determined,  eq.  (3)  is  used.  ^incQ  dl^rd6,x  =  r  cos  d^ 
and  l  =  ra. 


ra 


/a/2 
rd( 
a/ 2 


rdd-r  cos  d,     or 


._     2r    .    a 

x  =  —  sm  — . 
a         2 


For  a  semicircular  arc,  a  =  180°,  and  x  =  2r/n  =  o.6^'jr. 
2.  Locate  the  centroid  of  a  circular  arc  of  90°. 
Ans,  Distances  from  OA  and  OB  equal  2r/7r. 


3.  Locate  the  centroid  of  a  circular  sector. 

Solutions:  Evidently  the  centroid  is  on  OC  (fig.  53),  i.e., 
57  =  0.     To  determine  x,  eq.  (2)  is  used, 
(i)  If  the  elementary  area  be  chosen  so  that  dA=pdddp,  the 


70  CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III. 

X  coordinate  of  the  centroid  of  the  element  is  p  cos  6]  and  since 

\r'^a-x=   I      I  p^  cos  ddpdd,     or     ::k:  = -^  sin  — . 

Jo     ^-a/2  3a  2 

(2)  If  the  sector  is  considered  as  made  up  of  elementary  trian- 
gles, so  that  dA  =hr'^ddy  the  x  coordinate  of  the  centroid  of  any 
element  is  |r  cos  6.     Eq.  (2)  becomes 

/'a/2  _     Ar        a 

cos  Odd,    or     x  =  — sin—. 
a/2  3«         2 

(3)  The  student  should  write  the  expression  for  Ic,  choosing  the 
elementary  area  so  that  dA  =  dx  dy. 

For  a  semicircular  area,  «=  180°,  and  ^  =  4r/37r  =  o.424r. 

4.  Locate  the  centroid  of  a  circular  quadrant. 
Ans.  Distance  from  either  straight  side  is  4^/3^- 

5.  Show  by  the  integration  method  that  the  centroid  of  a 
triangle  is  one-third  the  altitude  from  the  base. 

(Suggestion:  Take  the  origin  of  coordinates  at  the  vertex 
and  consider  the  triangle  as  consisting  of  elementary  strips 
parallel  to  the  base.) 

6.  Locate  the  centroid  of  a  symmetrical  parabolic  segment 
whose  altitude  is  a. 

Ans.  It  is  on  the  axis  of  the  parabola  distant  fa  from  the 
vertex. 

7.  Locate  the  centroids  of  the  halves  into  which  the  axis 
divides  the  segment  described  in  the  preceding  example. 

84.  Surfaces  of  Revolution. — General  formula  (2)  of  the  pre- 
ceding article  is  used;  it  will  be  advantageous  to  select  the 
elementary  area  in  a  certain  way,  namely,  the 
area  described  by  an  elementary  part  of  the 
generating  curve.  Let  the  x  axis  be  taken 
coincident  with  the  axis  of  revolution,  fig.  54; 
then  the  area  described  by  a  part  of  the  gen- 
erating curve  whose  length  is  ds  is  27tyds.  The 
coordinates  x,  y,  and  z  of  the  centroid  of  this 
Fig  V-i"""*  ^^^^  ^^^  evidently  x,  o,  and  0  respectively; 
hence 

Ax  =  27:  I  yds- X,     or    x=-t-  /  xyds,     and     57  =  2=0. 


§111.]  CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES,  n 

A  stands  for  the  area  of  the  surface  of  revolution,  and 

A  =  27: 1  yds. 

The  limits  of  integration  must  be  assigned  so  that  each  element 
ds  is  represented  in  the  integrations. 

EXAMPLES. 

I .  Locate  the  centroid  of  a  segment  of  a  spherical  surface. 

Solution:  The  segment  may  be  considered  as  generated  by 
the  revolution  of  a  circular  arc,  as  ^C  about  OC,  fig.  52.  Since 
x  =  r  cos  6,  y  =  r  sin  6,  and  ds  =  rdd, 


^  / 


cos  d  sin  6  dd 


i 


sin  d  dd 


=  -   I  +  cos  —   , 

2\  2/ 


2.  Show  by  the  integration  method  that  the  centroid  of  the 
surface  of  a  cone  is  one-third  of  the  altitude  from  the  base. 

85.  Solids  of  Revolution. — General  formula  (i),  art.  83,  is 
used;  it  will  be  advantageous  to  select  a 
certain  elementary  volume,  namely,  that 
generated  by  an  elementary  part  of  the  gen- 
erating plane  which  is  included  between  two 
lines  perpendicular  to  the  axis  of  revolution. 
Thus  if,  in  fig.  55,  the  generating  plane  is  that  q 
bounded  by  the  solid  curve,  and  the  x  axis  is 
taken  coincident  with  the  axis  of  revolution, 

dV  =  ^{y^  —  y^)dx. 
Now  the  centroid  of  this  elementary  volume*  is 
in  the  x  axis,  and  its  x  coordinate  is  the  ^in 
the  figure;  hence 

Vx=7z  I  {y^-y^)dx'X,     or     x=^  I    {y^-y^)xdx, 
V  denotes  the  volume  of  the  solid  of  revolution,  and 

The  limits  of  integration  are  to  be  assigned  so  that  each  dV  is 
represented  in  the  integrations. 


Fig.  55. 


72 


CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III.' 


EXAMPLES. 

I.  Locate  the  centroid  of  a  segment  of  a  sphere.  i 

Solution :  The  segment  may  be  considered  as  generated  by  i 

the  revolution  of  one-half  of  a  circular  segment,  as  ACD  about  \ 

OC,  fig.  56.     Since  x  =  r  cos  d,  dx=—r  sin  6  dd.     Also  ^^j  =  r'sin;  ^  ! 

and  >'i  =  o,  therefore,    ^    ^  I 


x  = 


■a/, 


sin^^cos  ddd 


/:'■' 


sin^  d  dd 


^^ sin^  a/ 2 

*    2— 3  cos  a/2+cos^  a/2' 


2.  Locate  the  centroid  of  a  paraboloid  of  revolution  whose  \ 

altitude  is  a.  \ 

Ans.  It  is  in  the  axis  of  revolution  distant  fa  from  the  ' 

apex.  ] 

y        ^  y 


Fig.  56. 
3.  Locate  the  centroid  of  a  frustum  of  a  cone. 
Suggestion:    The  frustum  may  be  considered  as  generated 
by  the  revolution  of  the  shaded  trapezoid  (fig.  56)  about  Ox. 

86.  Theorems  of  Pappus  and  Guldinus. — L  The  area  of  the 
surface  of  revolution  generated  by  a  plane  purve  revolved  about 
an  axis  in  its  plane  equals  the  length  of  the  curve  times  the 
circumference  of  the  circle  described  by  its  centroid. 

Proof:  Let  A  denote  the  area  of  the  surface,  /  the  length  of 
the  curve,  and  y  the  ordinate  of  its  centroid  measured  from  the 
axis.     Then  (see  fig.  54) 


A 
hence 


=  27:  I  yd. 


s     and,  from  eq.  (3),  art.  S^,     y 


Lfyds; 


K  =  \'27ty (i) 


.  I 


§111.]  CENTROIDS  OF  SOLIDS,  SURFACESy  AND  LINES.  73 

?  II.  The  volume  of  a  solid  of  revolution  generated  by  a  plane 
figure  revolved  about  an  axis  in  its  plane  equals  the  area  of  the 
figure  times  the  circumference  of  the  circle  described  by  its 
centroid. 

Proof:  Let  V  denote  the  volume  of  the  solid,  A  the  area  of 
the  plane  figure,  and  y  the  ordinate  of  the  centroid  of  A  meas- 
ured from  the  axis.     Then  (see  fig.  55) 

and,  from  eq.  (2),  art.  ^t,, 

hence 

V  =  A.2;ry .     (2) 

Equations  (i)  and  (2)  are  available  for  con\puting  y  as  well 
as  ^  or  y  if  all  other  factors  in  the  equations  are  known.  It 
should  be  remembered  that  A  has  different  meanings  in  (i) 
and  (2), 

EXAMPLES. 

1.  Knowing  that  the  surface  of  a  sphere  is  4;rr^,  locate  the 
centroid  of  a  semicircular  arc. 

2.  Knowing  that  the  volume  of  a  sphere  is  |:rr^,  locate  the 
centroid  of  a  semicircle. 

3.  A  circle  of  radius  r  is  revolved  about  a  line  in  its  plane 
whose  distance  from  the  centre  is  a,  a  being  greater  than  r. 
Deduce  expressions  for  the  surface  area  and  volume  of  the  solid 
generated. 

4.  By  eqs.  (i)  and  (2),  deduce  formulas  for  the  surface  area 
and  volume  of  a  cone. 

5 .  Show  that  the  area  of  the  spherical  segment  generated  by 
revolving  the  arc  ACy  fig.  53,  about  Ox  is  27rr^(i  — cosa:/2). 

87.  Graphical  Determination  of  the  Centroid  of  a  Plane 
Figure. — I.  The  figure  can  be  subdivided  into  parts  whose 
areas  and  centroids  are  known.  Imagine  a  uniform  lamina 
shaped  like  the  figure  and  then  determine  its  centre  of  gravity 
by  the  method  outlined  in  art.  63,  p.  54.     As  an  illustration, 


73^ 


CENTRE  OF  GRAVITY  AND  CENTROID.         [Chap.  III.  i 


let  it  be  required  to  determine  the  centroid  of  the  shaded  part^ 
of  fig-  57(«)-  Imagining  the  figure  to  represent  a  uniform  ] 
lamina,  first    determine    the    action    line   of   the    resultant  of^ 


Fig.  57a. 


Fig.  s1^' 


the  weights  of  the  three  component  parts  of  the  lamina  act- 
ing through  their  respective  centres  of  gravity ;  these  weights 
are  proportional  to  the  areas  of  the  component  parts  of  the 
figure.  First  draw  a  force  polygon  for  the  forces,  as  ABCD, 
then  a  funicular  polygon,  as  oa,  oh,  oc,  od,  and  extend  the 
first  and  last  strings  oa  and  od;  then  a  line  through  their  inter- 
section parallel  to  the  forces  is  the  action  line  sought,  a(i.  Next, 
instead  of  turning  the  lamina  about  as  suggested  in  art.  63, 
imagine  the  direction  of  gravity  to  be  changed  so  that  a'h', 
b'c\  and  c'J' will  represent"  the  weights  of  the  parts  of  the  lam- 
ina; then  find  the  action  line  of  the  resultant  of  these  weights. 
A  second  force  polygon  is  unnecessary,  for  the  second  funicular 
polygon,  oV,  o'h' ,  o'c\  o'd' ,  can  be  constructed  from  the  first 
force  polygon,  o'a'  being  perpendicular  to  OA,  o'h'  perpendicular 
to  OB,  etc.  The  intersection  of  o'a'  and  o'd'  fixes  a'd' ,  which 
is  the  action  line  sought,  and  the  intersection  of  ad  and  a'd' 
is  the  centroid  of  the  entire  figure. 


I  §  III.]  CENTROIDS  OF  SOLIDS,  SURFACES,  AND  LINES.  73^^ 

In  this  case,  the  graphical  method  is  not  so  convenient  as 
the  algebraic,  but  in  the  following  case  it  is  generally  better. 

11.  The  figure  cannot  be  divided  into  simple  parts  whose 
areas  and  centroids  are  known. — The  method  requires  the 
use  of  a  planimeter  or  other  device  for  determming  the  area 
of  an  irregular  figure.  Let  abba  (fig.  57/^)  be  the  figure  whose 
centroid  is  to  be  determined,  OX  any  convenient  reference 
axis,  and  P  any  point  on  it.  (i)  Draw  O^X^  and  O^X^Sit  any 
convenient  distance  m  from  OX.  (2)  Draw  a  line  aa  parallel 
to  the  X  axis   through  the  figure  and  mark   its   intersections 

I  with    the   perimeter   a.      (3)  Project    aa    on    O^X,.      (4)  Join 

I  the  ends  of  the  projection  with  P  and  mark  the  intersections 
of  aa  with  the  joining  lines  a'.     (5)  Repeat  the  construction 

I  for  other  lines  like  da,  as  bb,  thus  locating  points  b' .  (6)  Draw 
a  smooth  curve  through  all  points,  as  a'6V,  etc.  (7)  Measure 
the  area  of  the  two  loops  *  (shaded  in  the  figure)  and  that  of 
the  given  figure.  (8)  Finally,  if  A  denotes  the  area  of  the 
given  figure,  ^/  and  A^'  the  areas  of  the  loops  on  the  positive 
and  negative  sides  of  the  x  axis,  and  y  the  ordinate  of  the  cen- 
troid, then 

I  y:=m{A,'-A,^)/A. 

1      Proof:  The  moment  of  the  area  of  the  given  figure  with 
•respect  to  the  x  axis  is 

P  'yA  =    /  yd  A  =   /      ywdy  =    /     ywdy  +    /      ywdy, 

»/  ^   J  —C2  J  —C2  Jo 

w  denoting  any  width  of  the  figure,  as  aa  or  bb.  If  w'  denotes 
the  width  of  the  loop  corresponding  to  w,  i.e.,  a' a'  or  b'b' ,  then, 
from  similar  triangles  in  the  figure,  it  follows  that 

^yjw'  =m/w,     or     yw  =  ±m'w\ 
the  positive  or  negative  sign  to  be  used  according  as  w  and  w' 
refer  to  widths  above  or  below  the  x  axis.-   Substituting  this 
value   of  yw  in  the  last  expression   for  the  moment,  we  have 

*'  'yA=m\   —  I      w'dy-\-    I      w'dy    . 

Now  the  first  integral  equals  A^Sind.  the  second  ^/ ;   hence,  etc. 
*  If  the  X  axis  does  not  cut  the  given  figure,  there  will  be  only  one  loop. 


CHAPTER    IV. 

i 
ATTRACTION  AND  STRESS*  -       I 

] 
§  L     Gravitation.  ] 

Every  body  in  the  universe  attracts  every  other  body,  the ' 
attraction  in  each  case  depending  upon  the  masses  of  the  two  ] 
bodies  and  their  distance  apart.  This  attraction  is  called  ] 
gravity,  gravitation,  and  gravitational  attraction. 

88.  Law  of  Gravitation. — Every  particle  attracts  every  other  ■ 
particle  with  a  force  which  is  proportional  to  the  product  of  the  "| 
masses  of  the  two  particles  directly  and  to  the  square  of  their^ 
distance  apart  inversely.  1 

The  law  may  be  stated  algebraically  thus :  let  F  denote  the  < 
attraction,  m' and  m"  the  masses,  and  r  the  distance;  then 

_    m'm"  _     ,  m'm"  j 

k  being  a  constant  whose  value  depends,  as  explained  below,  i 
upon  the  units  used  to  express  force,  mass,  and  distance.  ; 

Gravitation  Constant. — It  is  shown  in  art,  97  that  two  homo-  ; 
geneous  spheres  attract  each  other  as  though  the  mass  of  s 
each  were  concentrated  at  its  centre,  i.e.,  the  attraction  is  the  j 
same  as  that  between  two  particles  placed  at  the  centres  of  the  : 
spheres,  the  mass  of  each  being  equal  to  that  of  the  correspond-  1 
ing  sphere.  Hence  the  formula  for  the  attraction  between  par-  ] 
tides  applies  to  that  between  homogeneous  spheres,  r  denoting  : 

*  The  principles  of  mechanics  employed  in  this  chapter  are  mainly  ■ 
those  relating  to  composition  of  forces,  and  most  of  the  chapter  might  ' 
have  been  distributed  as  problems  on  composition  in  Chap.  II.  How-  ] 
ever,  as  it  all  relates  to  but  two  subjects,  it  is  convenient  to  treat  the  i 
matter  together.  I 

74  S 


§1.]  GRAVITATION.  75 

the  distance  between  their  centres ;  and  the  attraction  between 
two  spheres  of  unit  mass,  their  centres  being  unit  distance  apart, 

is  k  — ^  =k. 

Therefore,  the  numerical  value  of  k  equals  the  attraction 
between  two  homogeneous  spheres  of  unit  mass  whose  centres 
are  unit  distance  apart. 

The  numerical  value  of  the  gravitation  constant,  k,  hence 
depends  upon  the  units  employed  for  expressing  force,  mass, 
and  distance.  If  the  pound,  pound,  and  foot  respectively  be 
employed,  ^  =  3.31X10"",  and  for  C.G.S.  units  (see  art.  232) 
^  =  6.65X  io~*.  It  is  possible  to  choose  the  units  of  force,  mass, 
and  length  so  as  to  make  k=i',  two  of  them  may  be  chosen 
arbitrarily,  but  such  a  third  unit  is  an  uncommon  one,  and 
therefore  k  will  be  retained  in  the  formula. 

A  determination  of  the  gravitation  constant  involves  the 
measurement  of  the  attraction  between  known  masses  at  known 
distance  apart.  If  F  denotes  the  measured  force,  m'  and  m" 
the  masses  of  two  homogeneous  attracting  spheres,  and  r  the 
distance  between  their  centres,  then 

EXAMPLE. 

In  an  early  determination  of  the  gravitation  constant,  the 
attracting  spheres  were  of  lead,  two  and  twelve  inches  in  diam- 
eter. When  their  centres  were  nine  inches  apart,  the  attraction 
was  measured.     What  was  its  value  in  pounds  ? 

89.  Density. — By  density  of  a  body  is  meant  its  mass  per 
unit  volume.     We  will  denote  it  by  d. 

In  a  homogeneous  body ,  the  mass  of  a  unit  volume  is  the  same 
no  matter  where,  in  the  body,  the  volume  is  selected;  hence  the 
density  is  constant,  and  its  value  is  found  by  dividing  the  mass 
of  the  body  by  its  volume,  i.e.,  if  m  and  V  denote  mass  and 
volume  of  the  body  respectively, 


76  ATTRACTION  AND  STRESS.  [Chap.  IV.  i 

In  a  heterogeneous  body,  the  density  is  variable  and  the  ex- j 
pression  m/V  gives  the  average  deajsity;  and  if  Jm  denotes  thei 
mass  of  a  portion  whose  volume  is  J  F,  ^ 

-TY^  =  average  density  of  the  portion. 

If  the  small  portion,  as  its  volume  approaches  zero,  always : 
includes  a  point  P,  then  \ 

Awl 
lim.  -Ty.  =  density  at  P, 

4 

or,  if  5  denotes  density  at  P, 

dm  j 

dV-  \ 

go.  Attraction   at    a   Point    or    "Strength   of   Field."— By] 

attraction  at  a  point  due  to  any  body  is  meant  the  attraction! 
which  it  would  exert  upon  a  particle  of  unit  mass  placed  at  the ; 
point  or  upon  a  homogeneous  sphere  of  unit  mass  whose  centre  ] 
is  at  the  point.  This  is  also  known  as  strength  of  field,  the  termj 
field  being  a  contraction  of  "field  of  force,"  which  means  the] 
region  at  all  points  of  which  there  is  an  attraction.  1 

91.  Attractions  in  Some  Simple  Cases.  —  To  compute  the- 
attraction  at  a  point  it  is  necessary  to  compound  the  gravita-| 
tional  forces  exerted  by  all  the  particles  of  the  attracting  bodys 
upon  a  particle  of  unit  mass  placed  at  the  point.  This  system  | 
of  forces  is  concurrent,  and  may  be  coUinear,  coplanar  or  non-^ 
coplanar;  hence  to  determine  the  resultant  of  the  system  the^ 
methods  of  §§  2,  3,  or  6,  Chap.  II,  may  be  used.  .| 

92.  Attraction  at  a  Point  on  the  Produced  Axis  of  a  Straight 
Slender  Rod.— Let  AB  (fig.  58)  be  the  rod  and  P  the  point.     If  | 

P  A Q B 

I ^  m  I 

k — X- — ->  1  \ 

•< av ^  ^ 

Fig.  58.  *  ] 

a  denotes  the  area  of  the  cross-section  of  the  rod,  and  dm  the' 
mass  of  an  elementary  volume  such  as  that  represented  at  Q,: 
then  dm  =  dadx.     Such  an  element  may  be  considered  as  a  par-  i 


I.] 


GRAVITATION. 


77 


icle,  for  its  dimensions  are  negligible  in  comparison  with  x\ 
hence  the  attraction  due  to  it  at  P  is  kdadx/x^.  Denoting  the 
attraction  of  the  whole  rod  by  F,  then 

y^,  x^        V^i   ^2/     I  Vi    ^2/ 

m  and  /  denoting  mass  and  length  of  the  rod  respectively. 

93.  Attraction  at  any  Point  Due  to  a  Straight  Slender  Rod. — 

Let  AB  (fig.  59)  be  the  rod  and  P  the  point  c 

distant  from  the  rod.  If  a  denotes  the  cross- 
sectional  area  of  the  rod  and  dm  the  mass  of 
Ian  element  such  as  that  represented  at  Q, 
|:then  dm  =  dady.     Such  an  element  may  be 

considered  as  a  particle,  since  its  dimensions 

are  negligible   compared  with   r;    hence   the 

attraction  due  to  it  at  P  is  kdady/r"^. 

The  X-  and  ;v- components  of  this   attrac- 
rtion  are  respectively  Fig.  ^9. 


/ 


i 


kdady 


cos  6     and 


kdady  . 


sin  d. 


It  is  plain  from  the  figure  that  r  =  c/cos  d,  and  y  —  c  tan  6;  hence 
4y  =  cdd/cos^d  and  the  expressions  above  can  be  written 


kda 


cos  d  dd     and 


kda 


sin  6  dd. 


If  Fx  and  Fy  denote  the  x-  and  :v-components  of  the  attraction 
at  P, 


»nd 


_kda    n 
_kda    r» 


cos  ddd  = (sin/?  +  sin  a), 


sin  ddd  = (cos /?  — cos  a). 


If  F  denote  the  magnitude  of  the  resultant  attraction  and  (f>  its 
wangle  with  the  x  axis, 

a  — 3 
tan  ^  =  F^/Fa;  =  tan -,     or     (j)  =  (a—^)/2. 


78 


ATTRACTIOl^  AND  STRESS. 


[Chap.  IV. 


94.  Attraction  Due  to  a  Circular  Ring  at  a  Point  on  its  Axis  J 
—Let  the  ring  be  that  representefl  in  fig.  60  and  P  the  point  h 

distant  from  its  centre.     Imagine  the; 

ring   resolved    into    elements   as   that 

represented  at  Q,  and  call  the  mass  of  s 

one   dm.      Such    an   element   may  be' 

treated   as   a  particle,  for  its  dimen-; 

sions  are  negligible  compared  with  PQ.  • 

The  attraction  at  P  due   to  any  ele-! 

me'nt    is    kdm/{r^-{-b^).       It    is    plain' 

from  the  symmetry  that  the  algebraic 

sums  of  the   y-  and  2-components  of| 

^^^-  ^°-  the  attractions  of  all  the  elements  arei 

zero,  hence  the  attraction  at  P  due  to  the  ring  equals  the  sumj 

of  the  ^-components.     Let  ^  denote  the  angle  CPQ,  then  i 

the  :3C-component  of  the  attraction  of  an  element  is  ? 

kdm  ,  \ 

and  if  F  denotes  the  resultant  attraction, 
^         k  .    r  ,  km 


r^  +  b' 


cos 


(j)  I  dm 


r^  +  b' 


cos  ^, 


m  denoting  the  mass  of  the  ring. 

95.  Attraction  Due  to  a  Thin  Circular  Plate  at  any  Point  on 
its  Axis. — Let  AB  (fig.  61)  be  the  plate, 
a  its  radius,  /  its  thickness,  and  P  the 
point.  Imagine  the  plate  as  consisting 
of  rings,  or  hollow  cylinders,  whose 
inner  and  outer  radii  are  r  and  r  +  dr 
respectively.  The  mass  of  such  a  ring 
is  d27zrdr-t,  and  the  attraction  of  the 
ring  at  P  equals 


,   d2nrdr-t  , 

^     r2j^l,2    cos  <j> 


Fig.  61. 


(see  art.  94). 
then 


Let  F  denote  the  attraction  of  the  whole  plate, 


F  =  27:kdt 


£■ 


'S^lldr. 


+b 


§LJ  GRAVITATION, 

Since  r  =  h  tan  <j),  dr  =  bd<j)/cos^<j),  and 

F  =  2nkdt 


79 


t  I    sin^(i0  = 


m. 


2nkdt{i— cos  a)  =  27:k-^{i  —cos  a). 


The  expression  2;r(i— cos<^)  is  the  value  of  the  solid  angle* 
subtended  by  the  plate  at  P\  hence,  if  oj  denote  that  angle, 


77      u*^ 


For  a  point  very  near  the  plate  a  is  nearly  90°,  and 


m 


F  =  k2n-7ry  approximately. 

96.  Attraction  Due  to  a  Spherical  Shell  at  any  Point. — Let 
ABC  (fig.  62)  be  the  shell,  represented  by  a  diametral  section. 


Fig.  62. 


and  P  the  point.  Imagine  the  shell  resolved  into  elementary 
rings  which  are  cut  out  from  the  shell  by  cones  whose  common 
apex  and  axis  are  O  and  OP  respectively.  Two  of  the  ring  sur- 
faces then  are  conical  and  two  are  zones  of  the  outer  and  inner  sur- 


*  The  student  is  reminded  that  a  solid  angle  is  measured  by  the  ratio 
)f  the  area  of  that  part  of  a  sphere,  whose  centre  is  at  the  apex  of  the 
mgle,  which  is  included  within  the  surface  bounding  the  angle  to  the 
;quare  of  the  radius  of  the  sphere.  Thus  if  for  the  solid  angle  sub- 
;ended  by  the  plate  at  P  a  sphere  of  radius  R  be  used,  the  cone  bounding 
;he  angle  cuts  from  the  sphere  a  segment  whose  area  is  2r.R*{i  —cos a), 
«ee  ex.  5,  art.  86;  hence  the  soHd  angle  equals  27r(i  —cos  a). 


OF    THE 


So  ATTRACTION  AND  STRESS. 


[Chap.  IV. 


faces  of  the  shell.  Let  dA  denote  the  area  of  the  inner  zonej 
and  /  the  thickness  of  the  shell;*  then  the  mass  of  the  ring  is 
ddA'ta.n6. 

the  attraction  of  the  ring  at  P  = ^-^  qos  (j),  \ 

(see  art.  94).     Let  F  denote  the  attraction  of  the  shell,  then      1 


F  =  kdtp-^dA. 


r 

the  limits  being  chosen  so  that  all  rings  of  the  shell  are  included.] 

To  integrate,  we  will  express  dA  and  cos  (ji  in  terms  of  rj 

From  the  figure,  i 

i 

dA  =  2na  sin  d '  add  =  2710?  sin  d  dd ,  i 

and  r^  =  a^-\-b^  —  2abcosd,     or     rdr  =  absm6dd;  • 

hence  , 

7  A  CL      ,  \ 

aA  =  27:  -y-r  dr.  1 

0 

Since  a^=r^  + 6^  —  27*6  cos  ^, 

cos  (j)  =  {r^  +  b^  —  a^)/2br. 
Finally, 

/ 

The  integral  equals  /  dr  +  {b^  —  a^)  I  —  = 

The  limits  of  the  integration  depend  on  whether  the  point  P\\ 
is  within  or  without  the  shell. 

(a)  When  the  point  is  external, 

7rfetorr2  +  a2-62-i6+a      kd^naH      km 


i^  =  -p-   /  -. dr. 


rr^  +  a^-6^~1^+^_ 


F  =  ' 

'"    '  '  '"  b^  ' 

m  denoting  the  mass  of  the  shell.  The  final  expression  shows 
that  the  attraction  at  a  point  outside  of  a  homogeneous  shell  is 
the  same  as  though  its  mass  were  concentrated  at  its  centre. 

When  the  point  is  on    the    surface,   b  =  a  and  F  =  k^K-^^ 
A  denoting  the  area  of  the  surface  of  the  shell. 


i 


§IL]  ELECTRIC  AND  MAGNETIC  ATTRACTIONS,  8i 

(6)  When  the  point  is  internal, 


rr2  +  a2_52^-|a- 


=  o. 

■b 


97.  Attraction  at  a  Point  Due  to  a  Sphere. — If  the  sphere 
may  be  resolved  into  homogeneous  shells,  the  results  of  the 
preceding  article  may  be  employed. 

For  exterior  points,  the  attraction  due  to  each  shell  is  the 
same  as  though  its  mass  were  concentrated  at  the  centre  and 
hence  the  attraction  due  to  the  sphere  is  the  same  as  though  its 
mass  were  concentrated  at  the  centre.  If  m  denotes  the  mass 
of  the  sphere  and  h  the  distance  of  any  exterior  point  from  the 
centre,  the  attraction  at  that  point  is  km/h^,  i.e.,  the  attraction 
varies  inversely  as  the  square  of  the  distance  from  the  centre. 

For  interior  points,  the  attraction  due  to  each  shell  which 
includes  the  point  is  zero,  and  the  attractions  of  all  shells  to 
which  the  point  is  external  are  the  same  as  though  their  masses 
were  concentrated  at  the  centre  of  the  sphere;  hence  only  the 
part  of  the  sphere  which  is  nearer  the  centre  than  the  point  is, 
exerts  an  attraction  at  that  point,  and  it  is  the  same  as  though 
the  mass  of  that  part  were  concentrated  at  the  centre.  If  mf 
denote  the  mass  of  that  part  and  h  the  distance  of  the  point  from 
the  centre,  the  attraction  at  that  point  is  km' /h"^. 

If  the  sphere  is  homogeneous,  w'  =  ^|7r6^;  hence  the  attrac- 
tion at  the  point  is  k^noh,  i.e..  it  varies  directly  as  the  distance 
from  the  centre. 

§  II.     Electric  and  Magnetic  Attractions. 

Electrified  bodies  attract  or  repel  each  other,  and  so  do 
magnetized  ones.  A  study  of  these  forces  does  not  fall  within 
the  scope  of  this  book,  but  their  laws  are  so  similar  to  that  of 
gravitation  that  several  important  propositions  relating  to 
electric  and  magnetic  attractions  can  be  readily  deduced  from 
the  results  derived  on  gravitation. 

98.  Laws  of  Electro-Static  and  Magnetic  Forces. — (i)  Bodies 
similarly  electrified  or  magnetized  repel,  and  those  dissimilarly 
electrified  or  magnetized  attract,  each  other. 


„1 

i 

^2  ATTRACTION  AND  STRESS.  [Chap.  IV.  \ 

\ 
This  law  points  out   a  difference  between  electric  or  mag-  \ 

netic  force  and  gravitation;  the  latter  is  always  attractive.       \ 

(2)  The  force,  attraction  or  repulsion,  between  two  bodies  \ 

electrified  or  magnetized  is  proportional  directly  to  the  product  \ 

of  the  quantities  of  electricity  or  magnetism  and  inversely  to  \ 

the  square  of  the  distance  between  them.  I 

In  order  that  the   distance  between   the   bodies  may  be   defi- 1 

nite  their   dimensions   must   be  negligible   compared  with  the  \ 

distance  between  any  two  points  of  the  bodies.*  | 

The  second  law  may  be  stated  algebraically;  thus,  let  F\ 

denote  the  force,  q'  and  q"  the  quantities,  and  r  the  distance,  | 

then  I 

Fo^if,    or    F  =  kif.   ......     (.)  i 

k  being  a  constant  corresponding  to  the  gravitation  constant,  j 
It  is  customary  to  employ  units  of  force,  quantity,  and  dis-  i 
tance  which  make  k  equal  to  unity;  then  | 

^      q'q"  ^ 

i 

99.  Strength  of  Field. — By  strength  of  field  at  any  point  \ 
due  to  any  quantity  of  electricity  or  magnetism  is  meant  the  1 
force  which  that  quantity  would  exert  upon  a  unit  quantity  | 
of  electricity  or  magnetism  concentrated  at  that  point.  It  \ 
is  assumed  that  the  unit  quantity  would  not  affect  the  field.        j 

100.  Analogy  between  Electrical  or  Magnetic  and  Gravita-  \ 
tional  Attractions. — Comparing  the  equation  of  art.  87  and  (i)  ; 
of  art.  98,  it  is  seen  that  they  are  perfectly  similar;  hence  the  \ 
strength  of  field  due  to  any  distribution  of  electricity  or  mag-  | 
netism  may  be  found  by  analogy  from  the  gravitational  at-  -i 
traction  due  to  a  body  the  distribution  of  whose  mass  is  simi-  1 
lar  to  that  of  the  electricity  or  magnetism.  \ 

10 1.  Strengths  of  Field  Due  to  Some  Simple  Distributions  j 

of   Electricity' and  Magnetism.  —  I.    Electrified   Circular   Plate.  ^ 

___  .  _, 

*  These  statements  of  the  laws  are  somewhat  loose,  but  they  serve  the    ;^ 

present  purpose  better  than  the  usual  forms.     The  reader  is  assumed  tc    '''. 

have  at  least  an  elementary  knowledge  of  the  phenomena  of  these  forces.     • 

.^ 


§  II.]  ELECTRIC  AND  MAGNETIC  ATTRACTIONS,  83 

— The  electricity  resides  at  the  surface,  and  is  nearly  uni- 
formly distributed,  being  more  "dense"  at  the  edges  than 
elsewhere  on  the  surface.  If  we  regard  it  uniformly  distributed, 
the  expression  for  the  strength  of  field  at  any  point  on  the 
axis  of  the  plate  due  to  the  electricity  on  either  face  may  be 
deduced  from  the  expression  for  the  gravitational  attraction 
due  to  a  homogeneous  thin  plate  at  any  point  of  its  axis,  which 

is  k-^co  (art.  95)       For  strength  of  field  due  to  the  electricity 

on  one  side,  we  make  k=i,  and  substitute  q  for  w,  q  denoting 
the  quantity  of  electricity  on  the  side  considered.  If  F  de- 
notes the  strength  of  field,  then 

Now  q/A  is  the  quantity  of  electricity  per  unit  area,  and  is 
called  the  surface  density ;  if  it  be  denoted  by  p, 

F  =  pco •     (i) 

For  a  point  at  the  centre  of  the  face,  (0  =  211,  and 

F  =  2np (2) 

If  the  plate  be  very  thin,  so  that  its  thickness  is  negligible 
icompared  to  the  distance  between  plate  and  the  point  of  the 
axis  considered,  the  strength  due  to  electricity  on  both  sides 
is  called  strength  due  to  the  plate,  and  its  value  is  qco/A,  q 
denoting  quantity  on  both  faces.  For  such  thin  plates  q/A, 
the  quantity  of  electricity  on  both  sides  per  unit  area  of  one 
side,  is  called  surface  density.  With  this  meaning  of  q/A,  or 
(0,  equations  (i)  and  (2)  may  be  used  to  compute  strengths 
due  to  thin  plates. 

II.  Electrified  Sphere. — The  electricity  resides  at  the  sur- 
face and  is  uniformly  distributed  over  it;  hence  the  expres- 
sion for  the  strength  of  field  due  to  an  electrified  sphere  may 
be  deduced  from  that  for  the  gravitational  attraction  due  to 
a  homogeneous  spherical  shell. 


84  ATTRACTION  AND  STRESS,  [Chap.  IV. 

(a)  For  external  points,  the  expression  is  km/h"^  (art.  96). 
For  strength  due  to  an  electrified  sphere  we  make  k=i,  and 
substitute  q  for  m,  q  denoting  the  quantity  of  electricity.  Then 
if  F  denote  strength, 

A  denoting  area  of  the  surface  of  the  sphere  and  r  its  radius. 
Now  q/A  is  the  surface  density;  if  it  be  denoted  by  p, 

^==-srp (3) 

For  a  point  at  the  surface  b  =  r,  and 

F  =  4^p (4) 

(b)  For  internal  points,  the  gravitational  attraction  due 
to  a  shell  is  zero,  and  hence  the  strength  of  field  due  to  an  elec- 
trified sphere  at  an  internal  point  is  also  zero. 

III.  Magnetic  Shell. — This  is  a  magnetized  plate,  the  mag- 
netism on  one  side  being  "north-seeking"  and  that  on  the 
other  side  "south-seeking."  The  expression  for  strength  of 
field  due  to  the  magnetism  on  one  side  of  a  circular  plate  at  a 
point  on  its  axis  may  be  deduced  from  the  expression  for  the 
gravitational  attraction  due  to  a  homogeneous  thin  circular  plate. 

That  expression  is  k'.oj  (art.   95).     For  strength  of  field  due 
A 

to  the  magnetism  on  one  side  we   make  k  =  i,  and   substitute 

q  for  m,  q  denoting  the  quantity  of  magnetism.     Then  if  F 

denote  strength, 

Now  q/A  is  the  quantity  of  magnetism  on  one  side  per  unit 
area,  and  is  called  the  surface  density  of  m.agnetism;  if  it  is 
denoted  by  ^o, 

F=p(o (S) 

For  a  point  at  the  surface  of  the  shell  (0  =  27:,  and 

F  =  2np.     .     ,     , (6) 


§IIL]  STRESS.  85 

§  III.     Stress. 

102.  Stress  Defined. — The  term  stress  is  variously  defined. 
Some  writers  mean  by  it  the  forces  which  any  two  bodies  or 
two  parts  of  a  body  exert  upon  each  other,  it  being  then  a 
term  which  refers  to  any  "action"  and  its  "reaction"  (art,  6). 
For  example,  such  writers  designate  as  a  stress  the  forces  which 
the  earth  and  sun  exert  upon  each  other,  the  forces  which  the 
upper  and  lower  halves  of  a  monument  exert  upon  each  other, 
etc. 

Most  engineers,  however,  use  the  term  in  a  narrower  sense, 
meaning  by  it  the  force  which  one  part  of  a  body  exerts  on  an 
adjacent  part  at  the  surface  of  contact  of  the  parts.  Such 
engineers  designate  as  stresses  the  force  which  the  upper  or 
lower  half  of  a  monument  exerts  upon  the  other  half,  the  force 
which  either  half  of  a  stretched  string  exerts  on  the  other  half, 
etc. 

We  will  use  the  term  in  the  engineer's  sense  slightly  ex- 
tended and  define  it  thus:  Stress  is  any  force  whose  place  of 
application  is  a  surface.  The  force  may  be  exerted  between 
parts  of  one  body  or  between  two  different  bodies  which  are 
in  contact,  a  part  or  all  of  the  contact  surface  being  the  place 
of  application  of  the  force. 

103.  Units  for  Expressing  Stress. — Since  a  stress  is  a  force, 
it  must  be  expressed  in  force  units,  the  pound  force,  the  kilo- 
gram force,  etc. 

104.  Classification  of  Stresses.— If  the  parts  of  a  stress  on 
all  equal  small  portions  of  its  place  of  application  are  equal, 
the  stress  is  said  to  be  uniform  as  to  distribution;  if  otherwise, 
non-uniform.  If  the  action  lines  of  the  resultant  forces  on 
all  the  small  portions  are  parallel,  the  stress  is  said  to  be  uni- 
form as  to  direction;  if  otherwise,  non-tiniform. 

If  the  action  lines  of  the  resultants  are  all  normal  or  tan- 
gential to  the  surface  of  application ,  the  stress  is  called  simple; 
if  otherwise,  complex.  Stresses  are,  in  general,  complex,  but 
it  is  possible  to  describe  any  complex  stress  in  terms  of  simple 
stresses;  only  such  are  discussed  herein. 

Simple  stresses  may  be  classified  into  normal  and  tangen- 


S6  ATTRACTION  AND  STRESS.  [Chap.  IV. 

iial  stresses  according  as  the  action  lines  of  all  the  forces 
on  the  small  portions  of  the  placef  of  application  are  normal 
or  tangential  to  the  surface  to  which  the  stress  is  applied. 
Normal  stresses  are  subdivided  into  pressures,  or  compressions, 
and  tensions  according  as  the  force,  or  stress,  acts  toward  or 
away  from  the  place  of  application.  A  tangential  stress  is 
also  called  a  shear. 

The  classification  may  be  presented  thus: 

(normal...  i  f  ^^sure 
Simple  stresses  ]  ( tension 

( tangential . .  shear 

105.  Description  of  a  Simple  Stress. — This  requires  a  state- 
ment as  to  its  kind  (pressure,  tension,  or  shear),  and  as  to  the 
manner  of  its  distribution.  The  distribution  is  described  by 
a  statement,  for  each  point  of  the  place  of  application,  of  the 
value  of  the 

106.  Intensity  of  Stress. — By  intensity  of  stress,  or  stress 
intensity,  at  any  point  of  the  place  of  application  of  the  stress 
is  meant  the  stress  per  unit  area  at  that  point. 

If  the  stress  is  uniform  as  to  distribution,  then  the  stress 
per  unit  area  is  the  same  at  all  points,  and  its  value  is  found 
by  dividing  the  stress  by  the  area  of  its  place  of  application. 
If  Fy  A,  and  p  denote  the  stress,  area,  and  intensity  respect- 
ively, 

P  =  F/A (I) 

If  the  stress  is  non-uniform  as  to  distribution,  then  the 
stress  per  unit  area  is  different  at  different  points,  and  the 
expression  F/A  gives  the  value  of  the  average  intensity.  Also 
if  AA  denotes  the  area  of  ajiy  part  of  the  place  of  application 
and  AF  is  the  value  of  the  stress  applied  to  that  part,  AF/AA 
is  the  value  of  the  average  intensity  of  that  part,  AF,  of  the 
whole  stress.  Now  if  A  A,  as  it  approaches  zero,  always  in- 
cludes a  point  P,  then  AF/AA  approaches,  in  general,  a  finite 
limit,  and  the  value  of  that  limit  is  the  intensity  of  stress  at 
P,  or 

dF 
P  =  dA (^) 


§iil; 


STRESS. 


87 


The  tinit  of  intensity  of  stress  depends  on  the  unit  used  for 
expressing  F  and  A.  If,  for  example,  the  pound  and  square 
inch  are  used  for  these  respectively*,  then  the  pound  per  square 
inch  is  the  corresponding  unit  for  stress  intensity. 

[Note:  What  is  herein  called  intensity  of  stress  is  very 
commonly  called  by  engineers  "unit  stress."  Strictly,  unit 
stress  is  the  general  name  for  the  units  employed  for  express- 
ing stresses,  the  pound,  kilogram,  etc.,  and  the  engineer's 
usage  is  avoided  in  this  book  as  being  confusing  to  the  student. 
Once  thoroughly  familiar  with  the  quantities  involved,  he  may 
safely  adopt  the  engineer's  term.] 

107.  Graphical   Representation    of   a   Simple    Stress.* — 

(a)  Normal  Stress. — Let  abed,  in  the  xy  plane  (fig.  63),  be 
the  place  of  application  of  the  stress,  and  imagine  ordinates 
erected  at  all  points  of  it  propor- 
tional to  the  intensities  of  stress  at 
the  points;  thus,  if  p  denote  the  in- 
tensity at  P,  z  the  ordinate,  and  k 
any  constant  (the  scale  number), 

p  =  kz,  or  z  =  p/k. 
Then  the  volume  of  the  solid  defined 
by  those  ordinates  represents  the 
stress,  for  its  altitude  at  any  point  represents  the  intensity 
there,  and  the  volume  represents  the  value  of  stress,  as  can 
be  shown,  thus: 

the  stress ,  or  F  =  JpdA  =  kfzdA , 

and  the  volume,  or  V=  I zdA ; 


hence 


F  =  kV,     or     V  =  F/k, 


If  the  stress  is  partly  tensile  and  partly  pressural,  the  ordi- 
nates corresponding  to  tensions  and  pressures  are  drawn  from 
the  plane  in  opposite  directions.     Then  p  is  regarded  as  posi- 


*  In  this  and  the  following  articles  it  is  assumed  that  the  place  of 
application  of  the  stress  is  plane,  but  some  of  the  results  are  not  restricted 
to  such  cases. 


88  ATTRACTION  AND  STRESS.  [Chap.  IV. 

tive  for  one  kind  and  negative  for  the  other  kind  of  stress. 
(See  fig.  64.)  r 

(b)  Tangential  Stress. — A  tangential  stress  which  is  uni- 
form as  to  direction  may  also  be  represented  by  the  method 
described  above. 

108.  Centre  of  Stress. — Any  stress  which  is  uniform  as  to 
direction  can  be  conceived  as  a  system  of  parallel  forces  having 
definite  application  points.  Thus  imagine  the  place  of  applica- 
tion of  such  a  stress  divided  into  elementary  parts,  and  let  dA 
denote  the  area  of  any  element  and  p  the  intensity  of  stress  at 
that  element;  then  the  stress  or  force  on  that  element  is  pdA. 
Now  all  such  forces  as  pdA  make  up  a  system  of  parallel  forces 
and  their  application  points  are  definite  points  of  the  place 
of  application  of  the  stress. 

Since  a  system  of  parallel  forces  having  definite  applica- 
tion points  has  a  centroid  (art.  62),  therefore  a  stress  which  is 
uniform  as  to  direction  has  a  centroid,  or  a  centre  as  it  is  more 
commonly  called. 

The  centre  of  stress  is  in  the  plane  of  the  place  of  applica- 
tion, and  it  is  where  the  action  line  of  the  resultant  of  the  ele- 
mentary forces,  pdA,  pierces  that  plane.     Formulas  for  the 

position  of  the  centre  of  stress  in  the 
plane  may  be  deduced  from  those  for 
the  centroid  of  a  system  of  parallel 
forces  (art.  63).  Thus,  let  fig.  64  rep- 
resent a  stress  and  Xc  and  yc  the  coor- 
dinates of  the  centre  of  the  stress. 
'^'  Also,  let  X    and  y  denote   the    coor- 

dinates  of  the   application   point   of  the   force  pdA    on    any 
elementary  area  dA.     Then  . 


I  pdA  •  X  J  pdA 


^c  =  — t^ and      yc  = 


•y 


F  —      -'^  F       ' 

F  denoting  the  value  of  the  stress,  or    pdA. 

Proposition. — A  line  drawn  through  the  centroid  of  the 
solid  representing  a  stress  and  perpendicular  to  its  place  of 
application  passes  through  the  centre  of  the  stress. 


§111.] 


STRESS. 


89 


Proof:  According  to   the   preceding   article,  the   equations 
above  can  be  rewritten  thus: 


Xr  = 


fzdA'X 


and    yc  = 


JzdA-y 


V       ' 


hence 


/ 


dV'X 


fdV 


•y 


Xc= — y —     and    yc=-      y      . 

fdV'X  JdV-y 
Now t7 —  ^^d  y^ —  are  the  expressions  for  the  coordi- 
nates of  the  centroid  of  the  solid  whose  volume  is  V  (art.  83) ; 
hence  the  corresponding  coordinates  {x  and  y)  of  the  centre 
of  stress  and  the  centroid  of  the  solid  representing  the  stress 
are  equal.  It  follows  that  the  line  joining  those  two  points 
is  parallel  to  the  z  axis,  hence,  etc. 

109.  A  Uniformly  Varying  Normal  Stress.-^ By  this  is 
meant  one  whose  intensity  at  any  point  is  proportional  to  the 
distance  of  that  point  from  some  straight  line  in  the  plane  of 
the  place  of  application  of  the  stress.  The  straight  line  is 
called  a  neutral  axis,  or  zero  line.  A  familiar  example  of  such 
a  stress  is  the  pressure  of  a  liquid  which  is  at  rest  upon  an  im- 
mersed flat  surface  which  is  not  horizontal,  and  an  important 
case  is  the  ** fibre  stress"  on  any  cross-section  of  a  moderately 
loaded  beam. 


Fig.  65  represents  three  such  stresses;  the  places  of  appli- 
cation are  abed,  and  the  zero  lines  are  coincident  with  the  y 
axes.     In  (a)  the  stress  is  part  tension,  part  pressure;  in  (6) 


90  ATTRACTION  AND  STRESS.  [Chap.  IV. 

and  {c)  it  is  all  of  one  kind.  In  any  case,  the  law  of  variation 
can  be  expressed  thus:  '■ 

p  =  ax, 

p  being  the  intensity  at  a  point  whose  distance  from  the  zero 
line  is  x  and  a  a  constant.  When  the  stress  consists  of  a  ten- 
sion and  a  pressure,  then  p  in  the  equation  must  be  regarded 
as  having  sign,  its  sign  being  the  same  as  that  of  x. 

When  x=i,  p  =  a\  hence  the  numerical  value  of  a  equals 
the  value  of  p  at  points  at  unit  distance  from  the  zero  line. 

The  value  of  the  stress. — The  stress  on  an  elementary  area 
is 

dF  =  pdA=axdA, 
hence 

F==afxdA (i) 

Since  JxdA  =  xA ,  A  being  the  area  of  the  place  of  application 

of  the  stress  and  x  the  x  coordinate  of  its  centroid, 

F  =  axA (2) 

Let  p  denote  the  intensity  of  stress  at  the  centroid  of  the  place 
of  application,  then 

p  =  ax     and     F  =  pA (3) 

Proposition. — The  intensity  of  a  uniformly  var}4ng  stress 
at  the  centroid  of  its  place  of  application  and  the  average  in- 
tensity of  the  stress  are  equal. 

Proof:  From  equation  (3),  p  =  F/A;  since  F/A  is  the  value 
of  the  average  intensity  (art.  106),  the  proposition  is  proved 
by  that  equation. 

no.  Centre  of  a  Uniformly  Varying  Stress. — Substituting 
values  of  p  and  F  for  a  uniformly  varying  stress  in  the  first 
expressions  for  the  Xc  and  yc  in  art.  108,  we  have 

af(xdA)x      Jx^dA 

af{xdA)y       JxydA 
^''^      a  Ax        ""     Ax     ' 


§111.]  STRESS.  91 

Ax  is  the  moment  of  the  area  of  the  place  of  application  of  the 
stress  with  respect  to  the  zero  line.  The  integrals  in  the  ex- 
pressions for  Xc  and  yc  are  of  great  importance  and  have  special 
names;  they  are  discussed  in  App.  D,  where  their  values  for 
many  forms  of  surfaces  are  deduced. 

EXAMPLES. 

1.  Where  is  the  centre  of  a  stress  whose  place  of  applica- 
tion is  a  rectangle,  the  intensity  at  any  point  being  propor- 
tional to  its  distance  from  one  side? 

2.  Where  is  the  centre  of  a  stress  whose  place  of  applica- 
tion is  a  semicircle,  the  intensity  at  any  point  being  propor* 
tional  to  its  distance  from  the  diameter? 

111.  Moment  of  a  Uniformly  Varying.  Normal  Stress. — The 
stress  on  an  element  of  area  dA  whose  ordinate  is  x,  is  axdA, 
and  its  moments  with  respect  to  the  x  and  y  axes  are  respect- 
ively 

{axdA)y     and     (axdA)yx. 

If  Mx  and  My  denote  the  moments  of  the  entire  stress  about 
the  X  and  y  axes  respectively, 

Mx  =  afxydA 

aiid 

My^afx^dA. 

112.  Position  of  the  Neutral  Axis  when  the  Stress  is  Part 
Tensile  and  Part  Pressural  and  the  Two  Parts  are  Equal. — The 
algebraic  sum  of  the  elementary  stresses  is  zero;  hence 

axA=o,     or     x  =  o, 

i.e.,  the  zero  line  contains  the  centroid  of  the  surface  of  appli- 
cation of  the  stress. 

In  a  beam,  as  ordinarily  loaded,  there  is  on  each  crofs- 
section  a  stress  of  the  kind  described  in  the  title  to  this  article, 
and  the  result  above  deduced  is  of  great  practical  importance. 
It  will  bear  another  deduction ;  it  differs  but  slightly  from  that 
given  above.     Let  A^  and  A"  be  the  areas  of  the  parts  of  the 


92  ATTRACTION  AND  STRESS.  [Chap.  IV. 

place  of  application  of  the  stress  which  sustains  tension  and 
pressure  respectively,  and  'x'  and  *X"  the  distances  (not  coor- 
dinates) of  their  centroids  from  the  zero  line.     Then 

the  value  of  the  tension  is  ax'A'\ 
**       *'      *'    *'    pressure  is  a3t''A"  (art.  1*09). 

Since  these  are  equal,  x'A'  =  x"A" ,  that  is,  the  moments  of 
Iho  two  parts  of  the  entire  area  A  about  the  yz  plane  are  equal; 
hence  that  plane  contains  the  centroid  of  the  whole  area  (art. 
71).  But  the  centroid  is  also  in  the  xy  plane,  hence  it  is  in 
the  y  axis,  or  the  neutral  axis. 


CHAPTER    V. 
GENERAL  PRINCIPLES  OF  EQUILIBRIUM. 

§  I.     Preliminary. 

113.  Definitions. — An  external  force  is  one  exerted  on  a 
body  by  some  other  body.  An  internal  force  is  one  exerted 
on  a  part  of  a  body  by  some  other  part.  The  same  force  may 
be  classed  either  as  external  or  internal,  depending  upon  the 
point  of  view.  For  example,  consider  a  block  resting  upon  a 
table;  the  two  bodies  exert  forces,  or  pressures  upon  each 
other.  If  the  block  and  table  are  considered  as  one  body,  then 
both  of  these  forces  are  internal,  but  if  they  are  considered  as 
two  separate  bodies,  then  each  of  the  forces  is  an  external  one. 
Again,  the  upper  and  lower  halves  of  the  block  exert  forces 
upon  each  other,  and  with  reference  to  the  whole  block  con- 
sidered as  one  body  each  is  an  internal  force;  but  with  refer- 
ence to  either  half  considered  as  a  body,  the  force  exerted 
upon  that  half  is  an  external  one. 

A  body  is  said  to  be  in  equilibrium  if  (i)  it  is  at  rest  or  (2) 
its  state  of  motion  is  unchanging.  The  first  is  the  important 
case  in  this  connection. 

The  system  of  external  forces  applied  to  a  body  which  is  in 
equilibrium  is  also  said  to  be  in  equilibrium. 

114.  General  Condition  of  Equilibrium  for  a  System  of  Forces 
Applied  to  a  Rigid  Body. — If  the  system  of  external  forces 
applied  to  a  rigid  body  is  in  equilibrium,  its  resultant  is  nil. 
For  the  system,  being  in  equilibrium,  produces  no  change  of 
motion,  and  therefore  its  resultant  would  produce  none;  hence 
the  resultant  must  be  nil. 

Conversely,  if  the  resultant  of  all  the  external  forces  applied 
to  a  rigid  body  is  nil,  that  system  of  forces  is  in  equilibrium. 

93 


94  GENERAL   PRINCIPLES   OF  EQUILIBRIUM.  [Chap.  V,  ^ 

For  the  resultant,  being  nil,  would  produce  no  motion,  there-  ! 
fore  its  equivalent,  the  system,  produces  none;  hence  it  is  in  ^ 
equilibrium.  i 

The  condition  fulfilled  by  a  system  in  equilibrium  and  the  J 
condition  to  insure  the  equilibrium  of  a  system  are  one  and  i 
the  same,  namely,  the  resultant  is  nil;  this  is  called,  therefore,! 
the  general  condition  of  equilibrium.  i 

115.  Equilibrium  of  a  System  of  Forces  Applied  to  a  Non-rigid  ' 
Body. — The  condition  stated  in  the  preceding  article  for  the] 
equilibrium  of  forces  applied  to  a  rigid  body  is,  as  shown,  both  1 
necessary  and  sufficient.  For  the  equilibrium  of  a  deform- ' 
able  body  it  is  necessary  that  the  resultant  of  the  external  '■ 
applied  forces  be  nil,  but  it  is  not  sufficient.  This  may  be  ] 
explained  by  illustration: 

Consider  the  water  in  a  cup;  the  system  of  external  forces] 
applied  to  it  consists  of  its  weight,  the  pressures  exerted  by  i 
the  cup,  and  the  air  pressure  on  top.  Now  if  this  same  sys- i 
tem  of  forces  could  be  applied  to  the  water  when  frozen,  it ; 
would  certainly  be  in  equilibrium  and  its  resultant  would  be  \ 
nil ;  hence  the  resultant  of  the  external  system  on  the  body  of  | 
water  is  nil.  But  any  forces  applied  to  the  body  of  ice  which  | 
with  the  weight  have  a  zero  resultant  will  maintain  its  equi-  \ 
librium;  thus,  a  single  vertical  force  equal  to  the  weight  and  j 
acting  through  the  centre  of  gravity  will  answer.  The  same  i 
force  applied  to  the  body  of  water  will  not  of  course  maintain  I 
its  equilibrium,  although  the  resultant  of  the  two  applied  j 
forces  is  nil.  Other  conditions  than  a  vanishing  resultant  \ 
therefore  are  necessary.  In  this  case,  pressure  must  be  ex- 1 
erted  on  the  entire  surface  of  the  body  of  water  except  the  top,  ] 
and  in  a  certain  manner.  .  ) 

Again,  consider  two  similar  boards  fastened  together  with  5 
one  nail  as  in  fig.  66(a)  and  two  fastened  with  several  nails  | 
as  in  fig.  66(6),  and  suppose  their  planes  vertical.  It  can  be  \ 
proved  experimentally  and  otherwise  that  the  equilibrium  \ 
of  the  first  two  boards  can  be  maintained  by  two  forces  such  I 
as  R^  and  i^".  Clearly,  the  same  two  forces  would  maintain  i 
the  equilibrium  of  the  second  two  boards ;  hence  the  resultant  of  \ 
the  system  R',  R'\  W\  and  W  is  zero.     But  any  forces  which  '\ 


» 


§n.] 


CONDITIONS  OF  EQUILIBRIUM. 


95 


together  with  W  and  W'  have  a  zero  resultant  would  main- 
tain the  equilibrium  of  the  boards  in  fig.  66(6);  thus,  a  single 
vertical  force,  as  R,  equal  to  W  and  W"  will  answer.  The  same 
force  applied  to  the  first  two  boards  will  not  of  course  main- 
tain their  equilibrium,  although  the  resultant  of  the  applied 


R'  R"\  R 

(«)  (b) 

Fig.  66. 
forces  is   zero.     Therefore,  other  conditions  than  a  vanishing 
resultant  are  necessary. 

Summing  up,  if  the  system  of  external  forces  applied  to  a 
deformable  body  is  in  equilibrium,  the  resultant  is  nil;  but 
the  converse  is  not  necessarily  true. 

§  II.     The   Conditions    of    Equilibrium    for   the   Various 
Classes  of  Force  Systems. 

ii6.  CoUinear  Forces. — The  algebraic  condition  of  equilib- 
rium is,  2¥  =  0f 
i.e.,  the  algebraic  sum  of  the  forces  equals  zero. 

The  graphical  condition  of  equilibrium  is  that  the  force 
polygon  for  the  forces  closes,  or,  the  vector  sum  of  the  forces 
equals  zero. 

For  if  2F  equals  zero  or  the  force  polygon  for  the  forces 
closes,  the  resultant  is  nil. 

117.  Coplanar  Concurrent  Non-Parallel  Forces. — The  algebraic 
conditions  of  equilibrium  may  be  expressed  in  various  ways : 
W  ^F,  =  o,     IFy  =  o; ,  .     (i) 

i.e.,  the  algebraic  sum  of   the  resolved  parts  of  the  forces 

along  two  lines,  x  and  y,  equals  zero. 
b)  i'F^  =  o,     IM  =  o;    ......     (2) 

i.e.,  the  algebraic  sum   of  the  resolved  parts  of  the  forces 

along   a   line   x   equals  zero  and  the  moment  sum  for  the 

forces  with  respect  to  a  point  equals  zero.     (The  direction 


f 


96  GENERAL  PRINCIPLES  OF  EQUILIBRIUM.  [Chap.  V! 


X  is  not  to  be  perpendicular  to  the  line  joining  the  common 
intersection  of  the  action  lines  o*f  the  forces  with  the  origin 
of  moments.) 
(c)  ^Ma=o,     l^b  =  0', (3) 

i.e.,  the  moment  sums  of  the  forces  for  each  of  two  origins 
equal  zero.       (The  origins  and  the  common  intersection  of 
the  action  lines  of  the  forces  are  not  to  be  coUinear.) 
For,  in  either  case  the  resultant  is  zero,  as  may  be  thus   ex- 
plained : 

(a)  According  to  art.  36,  the  resultant  R  of  the  system,  if 
there  is  one,  is  given  by 

R  =  {IFx  +  I^Fyy;   hence,   if  IF^  and  IFy   equal  zero,  R 
equals  zero. 

(b)  If  I  Ma  is  zero,  then  the  resultant,  if  there  is  one,  must 
pass  through  a  as  well  as  through  the  common  point 
of  the  action  lines  of  the  forces,  O.  If  the  angle  between 
Oa  and  the  x  axis  be  a,  Rx  =  R  cos  a  =  IFx]  and  since 
JFx  =  o  and  a  is  not  90°,  R  must  be  zero. 

{c)  As  before,  the  resultant,  if  there  is  one,  must  pass  through 

O  and  a;  but  if  J  Mi,  is  also  zero,  and  0,  a,  and  b  are  not  | 

collinear,  R  must  be  zero.                                                  *  i 

The   graphical   condition   of   equilibrium   is   that   the   force  j 

polygon  for  the  forces  closes,  i.e.,  the  vector  sum  of  the  forces  ; 

equals  zero.     For,  if  the  polygon  closes,  the  resultant  is  zero,  i 

see  art.  SS- 

118.  Special  Condition  of  Equilibrium  for  Three  Forces. — The  i 

following  form  of  the  algebraic  conditions  j 

is  often  more   convenient   of  application  1 

than  .the  conditions    given    in    the    pre-  \ 

ceding  article.  j 

Fi/sin  «!  =  Fj/sin  a^  =  Fg/sin  a^ ;  1 

i.e.,  each  force  is  proportional  to  the  sine  i 

of  the  angle  between  the  other  two  (see  \ 

fig.  67).     For,  from  the  force  polygon  for  1 

Fig.  67.               ^^^  forces,  which  closes  (fig.  676),  j 

Fi/sin  (i8o-ai)=F2/sin  (iSo-a.;) -=FJsm  (180-0:3),  or  | 

FJsin  ai  =  F2/sm  a2  =  F^/sm  a^.  | 

^  I 

i 


§11]  CONDITIONS  OF  EQUILIBRIUM.  97 

119.  Coplanar  Non-Concurrent  Parallel  Forces. — The  alge- 
braic conditions  of  equilibrium  may  be  expressed  in  several  ways : 

(a)  i'F  =  o,     2'M  =  o; (i) 

i.e.,  the  sum  of  the  forces  and  the  sum  of  the  moments 
of  the  forces  each  equal  zero. 

(b)  i'Ma  =  o,     i'Mb  =  o; (2) 

i.e.,  the  moment  sums  for  the  system  for  each  of  two 
origins  equal  zero,  (The  line  joining  the  origins  is  not  to 
be  parallel  to  the  forces.) 
For  in  either  case  the  resultant  is  zero,  as  may  be  thus  ex- 
plained: In  art.  42,  it  is  shown  that  the  resultant  of  such  a 
system  is  either  a  force  or  a  couple,  and  if  a  couple,  its  moment 
equals  the  algebraic  sum  of  the  moments  of  the  given  forces 
about  any  point. 

(a)  If  IF  equals  zero,  the  resultant  is  not  a  force,  and 
if  IM  equals  zero,  the  resultant  is  not  a  couple;  hence  the 
resultant  vanishes. 
(6)  If  IMa  equals  zero,  the  resultant  is  not  a  couple,  and 
the  resultant  force,  R,  (if  there  is  one,)  must  pass  through 
a.  But  if  I  Ml  equals  zero,  R  must  equal  zero,  or  pass 
through  h',  but  R  is  parallel  to  the  forces  of  the  system 
and  cannot  therefore  pass  through  a  and  h.  Hence  R 
equals  zero,  i.e.,  the  resultant  vanishes 

The  graphical  conditions  of  equilibrium  are  (i)  the  force 
polygon  and  (2)  the  funicular  polygon  for  the  forces  must  close. 
For,  if  the  force  polygon  closes,  the  resultant  is  not  a  force, 
and  if  the  funicular  polygon  closes,  it  is  not  a  couple  (see  art.  40) ; 
hence  the  resultant  vanishes. 

120.  Coplanar  Non-Concurrent  Non-Parallel  Forces.  —  The 
algebraic  conditions  of  equilibrium  are 

(a)  i'Fa:  =  o,     2'Fj,  =  o,  .i'M  =  o;      .     ,     .     .     (t) 

i.e.,  the  algebraic  sums  of  the  resolved  parts  of  all  the 
forces  along  each  of  two  lines  and  the  moment  sum  of  the 
forces  with  respect  to  any  origin  equal  zero. 


j 

98                      GENER/iL  PRINCIPLES  OF  EQUILIBRIUM.          [Chap.  V.  \ 

(b)  IF^  =  Oy     I^a  =  Oy     2'M5  =  o;    ....     (2)  I 

i.e.,  the  algebraic    sum    of   the    resolved    parts  of   all   the  1 

forces  along  any  line  and  the  moment  sums  for  the  system  ] 

with  respect  to  two  origins  equal  zero.     (The  direction  of  i 

resolution  is  not  to  be  perpendicular  to  the  line  joining  ] 

the  two  moment  origins.)  \ 

(c)  ^Ma  =  o,     2'M6  =  o,     JM.^o;        ...     (3)  j 

i.e.,  the    moment    sums    for   the    system    with    respect    to  | 

three   origins   equal  zero.     (The  three  moment  origins  are  \ 

not  to  be  coUinear.)  \ 

For,  in  either  case  the  resultant  is  nil,  which  may  be  proved  j 

as  follows:  It  is  shown  in  art.  47  that  the  resultant  of  a  system  ; 

of  this  kind  is  a  force  or  a  couple;  if  a  force,  its  magnitude  j 

equals   {^Fx  +^Fy  )"^,   and  if    a   couple,   its   moment   equals  i 

2M.                                                         -  I 

(a)  If  IFx  and  IFy  equal  zero,  the  resultant  is  not  a  force,  | 
and  if  JM  equals  zero, the  resultant  is  not  a  couple;  hence  • 
the  resultant  vanishes.  j 

(b)  If  IM  =  o,  the  resultant  is  not  a  couple,  and  the  result-  ' 
ant  force,  R,  (if  there  is  one)  must  pass  through  a  and  1 
b  in  order  to  make  2! Ma  and  JM^  equal  to  zero.  If  now  ; 
JFx(=Rx)  equals  zero,  R  must  equal  zero.  j 

(c)  If  IM  =  o,  the  resultant  if  there  is  one,  is  not  a  couple,  j 
but  a  force.  If  I  Ma,  2  Mi,  and  IMc  equal  zero,  this  re-  \ 
sultant  force  must  equal  zero,  or  else  pass  through  a,  b,  \ 
and  c.  Since  the  latter  case  is  impossible,  the  resultant  \ 
vanishes.                            ••  i 

The   graphical  conditions   of   equilibrium   are  (i)  the  force  \ 

polygon  and  (2)  the  funicular  polygon  must  close.     For,  if  the  \ 

force  polygon   closes,  the  resultant  is  not  a  force,  and  if  the  I 

funicular  polygon  closes,  it  is  not  a  couple;  hence  the  resultant  I 

vanishes  (see  art.  45).  ; 

121.  Condition   of  Equilibrium   for  Four  Forces. — The  re-  \ 

sultant  of  either  pair  balances  that  of  the  other  pair.     This  \ 


ill.]  CONDITIONS  OF  EQUILIBRIUM.  99 

special  condition  may  often  be  advantageously  employed  in 
graphical  solution^,  of  problems. 

122.  Non-Coplanar  Concurrent  Forces. — The  algebraic  con- 
ditions of  equilibrium  are: 

2'F^  =  o,     IFy  =  o,     I¥,  =  o; 

i.e.,  the  algebraic  sums  of  the  resolved  parts  of  all  the  forces 
along  three  directions,  not  coplanar,  equal  zero.     For,  in  art.  50 

it  is  shown  that  the  resultant  R  of  this  kind  of  a  system  is  given 

2  —  2     2  1 

by  R  =  {IFx  +^Fy  -{-IFg  )*;  hence  if  the  conditions  be  ful- 
filled, R  must  equal  zero. 

The  graphical  conditions  of  equilibrium  are  that  the  force 
polygons  for  two  projections  of  the  vectors  representing  the 
given  forces,  regarded  as  force  systems,  must  close.  For, 
the  closing  lines  of  such  polygons  are  the  projections  of  the 
resultant  of  the  given  forces,  and  if  the  force  polygons 
close,  the  projections  vanish  and  the  'resultant  is  zero  (seo 
art.  49). 

123.  Non-Coplanar  Non-Concurrent  Parallel  Forces. — The 
algebraic  conditions  of  equilibrium  are  : 

i'F  =  o,     2111^  =  0,     IMy  =  o; 

i.e.,  the  algebraic  sum  of  the  forces  and  their  moment  sums 
with  r,espect  to  two  axes  equal  zero.  (Neither  of  the  axes 
is  to  be  parallel  to  the  forces.)  For,  in  art.  54  it  is  shown 
that  the  resultant,  if  there  is  one,  is  a  force  or  a  couple.  If 
JF  is  zero,  the  resultant  is  not  a  force,  and  if  2Mx  and  ^My 
equal  zero,  the  resultant  is  not  a  couple;  hence  the  resultant 
vanishes. 

The  graphical  conditions  of  equilibrium  are  (i)  the  force 
polygon  for  the  forces  and  (2)  the  funicular  polygons  for  the 
projections  of  the  vectors  representing  the  given  forces,  regarded 
as  forces,  on  two  planes  must  close.  For,  the  resultant  of  the 
system,  if  there  is  one,  is  a  force  or  a  couple;  if  the  force  poly- 
gon closes,  the  resultant  is  not  a  force,  and  if  the  funicular 
polygons  close,  it  is  not  a  couple;  hence  the  resultant  vanishes 
(see  art.  52). 


loo  GENERAL  PRINCIPLES   OF  EQUILIBRIUM.  [Chap.  V.  ; 

124.  Non-Coplanar   Non-Concurrent   Non-Parallel   Forces. —  \ 

The  algebraic  conditions  of  equilibrium  are: 
i'F:,  =  o,     i'F^  =  o,     i'F,  =  o,     i'M^  =  o,     i'Mj,  =  o,     i'M,  =  o;   1 

i.e.,  the   algebraic   sums   of   the   resolved   parts   of  the   forces; 

along  three  lines  and  the  moment  sums  for  the  system  with  ] 

respect  to  three  axes  equal  zero.     (The  three  lines  must  not  be  ] 

coplanar,  nor  the  three  axes.)  \ 

For,  in  art.  55  it  is  shown  that  the  resultant  of  such  a  system  ^ 

is  a  force  and  a  couple,  and  in  art.  58  that  the  magnitude  of  ^ 

the   force   is    {JF^  +  IFy  +  IT^^)^ ,  and  the  moment  of  the  I 

couple  is  {IM^  +  ^^y  +  ^M^)^ ;  hence,  if  the  conditions  are  : 

fulfilled,  the  resultant  vanishes.  \ 

The   graphical   conditions   may   be    expressed  thus :    If   the  \ 

system  be  resolved  into  two  component  systems,  one  coplanar  : 

and  one  parallel,  the  forces  of  the  latter  being  perpendicular  | 

to  the  plane  of  the  former,  then  the  following  polygons  must  \ 

close :  ; 

the  force  polygon  for  the  coplanar  system,  \ 

*  *   funicular  polygon  for  the  coplanar  system,  \ 

"  force  polygon  for  the  non-coplanar  system,  ■ 

•*    funicular  polygons    for  the  projections  of  the  vec-  ; 

tors  representing  the  parallel  system    on   each  of  I 

two  non-parallel  planes.  \ 

For,  if  these  conditions  are  fulfilled,  the  resultants  of  the  compo-  \ 

nent  systems  are  zero  (see  arts.  45  and  52),  and  hence  the  re-  i 

suit  ant  of  the  system  itself  vanishes.  i 

125.  Special  Condition  of  Equilibrium  and  Summary. — 
Proposition. — If  three  forces  are  in  equilibrium,  they  must  | 

be  coplanar  and  concurrent  or  parallel.  1 

Proof:  The  resultant  of  the  first  two  forces  must  be  a  single  \ 
force,  since  it  must  balance  a  force,  the  third  one.  Since  the  j 
two  forces  ai^  their  resultant  are  coplanar  and  the  resultant  \ 
and  the  third  force  are  coUinear,  the  three  given  forces  are  co-  \ 
planar.  If  the  first  two  forces  are  parallel,  their  resultant  is  ; 
parallel  to  them,  and  hence  the  third  force  is  also;  if  the  first  ] 
two  forces  intersect,  their  resultant  passes  through  their  inter-  \ 
section,  and  hence  the  third  force  does  also.  ) 


§  II.]  CONDITIONS  OF  EQUILIBRIUM.  loi 

Algebraic  Conditions  of  Equilibrium. 

Coplanar  Systems : 

Collinear IF    =  o. 

Concurrent  Non-parallel IF^  =IFy  =  o, 

IF:,  =IM  =  o,  or 

IMa-IMb  =  o. 
Non-concurrent  Parallel IF    ==IM  =  o,  or 

IMa  =  IMb  =  o, 
Non-concurrent  Non-parallel  IF^  =IFy  =  IM  =  o, 

IF^  =IMa  =  IMi,  =  o,  or 

IMa  =  IMb==IMc  =  o. 
Non-coplanar  Systems : 

Concurrent. 2'Fx  =  IFy  =  IF^  =  o. 

Non-concurrent  Parallel IF     =IMx  =  IMy  =  o. 

Non-concurrent  Non-parallel .  IF^  =  IFy  =  IFg  =  o  and 

IM^  =  IMy  =  IM,  =  o. 

Graphical  Conditions  of  Equilibrium. 
Coplanar  Systems: 

Concurrent The  force  polygon  closes. 

Non-concurrent..  .The  force  and  funicular  polygons  close. 

Non-coplanar  Systems : 

Concurrent The    polygons    for   the  projections  of   the 

vectors  representing  the  system  on  any 
two  non-parallel  planes  close. 

Non-concurrent 

Parallel The  force  polygon  closes  and  the  funicular 

polygons  for  the  projections  of  the  vec- 
tors representing  the  system  on  two  non- 
parallel  planes  close. 

Non-concurrent 

Non-parallel. .  ..If  the  system  be  resolved  into  component 
systems,  one  coplanar  and  one  parallel, 
the  forces  of  the  latter  being  normal  to 
the  plane  of  the  former,  then  the  appro- 
priate conditions  stated  above  apply  to 
each  component  system. 


CHAPTER    VI. 
APPLICATIONS     OF     THE     PRINCIPLES     OF     EQUILIBRIUM. 

§  I.     Preliminary. 

126.  Nature  of  the  Problems. — The  problems  involved  in 
the  application  of  the  principles  of  equilibrium  are  usually  of 
this  kind:  a  system  of  forces  is  in  equilibrium  and  some  of 
them  are  partly  or  wholly  unknown ;  it  is  required  to  determine 
the  unknown  elements.  The  required  elements  may  be  the 
magnitudes,  directions,  or  action  lines  of  forces. 

127.  General  Method  of  Solution. — These  problems  can  be 
solved  by  two  methods,  the  algebraic  and  the  graphical.  The 
algebraic  method  is  to  write  the  appropriate  equations  of  equi- 
librium for  the  kind  of  a  force  system  under  consideration,  and 
then  to  solve  them  for  the  unknown  quantities.  This  process 
is  called  applying  the  algebraic  conditions  of  equilibrium.  The 
graphical  method  is  to  apply  the  appropriate  graphical  condi- 
tions of  equilibrium  for  the  kind  of  a  force  system  under  con- 
sideration. How  these  conditions  are  applied  is  explained  in 
the  solution  of  some  of  the  following  examples. 

Often  the  force  system  in  equilibrium  of  which  the  partly 
or  wholly  unknown  forces  are  a  part,  and  to  which  the  condi- 
tions of  equilibrium  are  to  be  applied,  is  not  specified.  In  such 
cases  the  system  may  be  recognized  by  directing  one's  attention 
to  a  body  which  is  in  equilibrium  and  with  reference  to  which 
some  or  all  the  unknown  forces  are  external.  All  the  external 
forces  applied  to  that  body  constitute  a  system  in  equilibrium. 

In  all  but  the  simplest  of  the  following  examples  the  student 
is  strongly  urged  to  make  a  sketch  representing  the  body  con- 
sidered and  the  external  forces  exerted  upon  it  before  apply- 
ing the  conditions  of  equilibrium.     He  will  be   aided  in  his 

102 


;  §  II.]  FLEXIBLE  CORDS.  103 

•  enumeration  of  the  external  forces  if  he  will  represent  first  the 
actions  through  distance  *  exerted  upon  the  body  by  other 
bodies  and  then  count  the  number  of  contacts  between  the 
body  under  consideration  and  other  bodies;  at  each  place  of 
contact  a  force  may  be  exerted  upon  the  body  (art.  7). 

§  II.     Flexible  Cords. 

128.  Definitions. — A  perfectly  flexible  cord  is  one  which 
may  be  bent   without  resistance.     Such   a  cord  is  ideal,  but 

=  some  cords  are  practically  perfectly  flexible;  such,  for  brevity, 
will  be  called  flexible,  other  cords  being  called  stiff.  A  heavy 
flexible  cord  if  unsupported  cannot  be  stretched  straight,  but 
will  sag  more  or  less,  depending  upon  the  applied  pulls.  The 
lighter  the  cord,  the  less  the  sag;  and,  if  the  weight  of  the  cord 
be  small  compared  with  the  pulls,  it  will  be  practically  straight. 
In  the  following,  cords  will  be  assumed  without  weight  except 
where  otherwise  stated. 

129.  Tension  in  a  Cord. — The  phrase  tension  in  a  cord  refers 
to  the  forces  which  two  parts  of  a  taut  cord  exert  upon  each 
other.  The  forces  are  equal  and  opposite  (see  art.  6).  By 
magnitude  of  the  tension  is  meant  the  magnitude  of  either  force. 

To  illustrate,  suppose  that  AB  (fig.  68)  is  a  flexible  cord 
without     weight    subjected    to     equal  a  c  b 

pulls  at  its  ends,  and  imagine  a  plane      p'  p" 

of  separation  at  any  place,  C,  between  Fig-  68. 

the  ends'  of  the  cord.  Since  the  part  AC  is  in  equilibrium,  a 
force  acts  upon  it  at  its  right  end  equal  and  opposite  to  P'; 
this  force  is  exerted  by  the  part  BC.  Similarly,  a  force  acts  on 
BC  at  its  left  end  equal  and  opposite  to  P" ;  this  force  is  ex- 
erted by  the  part  AC  These  two  equal  and  opposite  forces  at 
C  hold  the  parts  AC  and  BC  together.  The  magnitude  of  the 
tension  is  P'  (or  P")  no  matter  where  C  is  taken. 

If  the  pulls  P'  and  P"  are  unequal,  the  cord  will  not  be  in 
equilibrium,  and  the  magnitude  of  the  tension  varies  with 
the  section  C. 

*  The  only  actions  through  distance  considered  in  this  chapter  are 
the  weights  of  bodies. 


104    APPUCy4TIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 


EXAMPLES. 

1.  A  body  weighing  loo  lbs.  is  suspended  by  a  single  cord  \ 
which  is  deflected  from  the  vertical  by  a  horizontal  force  of  ; 
20  lbs.  applied  to  the  body.  How  great  are  the  deflection  and  ' 
the  tension  in  the  cord?  [ 

Solutions:    (i)    Algebraic.      Considering    the    forces  acting  J 

upon  the  body,  it  is  seen  that  they  are  three  in  number,  namely,  ? 

I  its  weight,   the  horizontal  force,  and  the  pull  of  i 

^1  the  string.       The  conditions  of  equilibrium,  with  I 

"•"V'l  axes  as  in  fig.  69,  are  (art.  125)  ? 

I  IFy=  —100 +  T  cos  d  =  o.  \ 

100  lbs.  Solving  these  it  is  found  that  ] 

Fig.  69.  \ 

r=  102  lbs.     and     ^=11°  19'.  \ 

(2)  Graphical.  The  condition  of  equilibrium  is  that  the  force  \ 
polygon  for  the  three  forces  must  close  (art.  125).  To  construct  1 
the  polygon,  the  wholly  known  forces  are  represented  first;  \ 
thus  AB  and  BC  (fig.  70)  represent  the  weight  and  the  hori-  \ 
zontal  pull  respectively,  scale  i  in.  =  100  lbs. 
The  closing  side  CA  then  represents  the 
magnitude  and  direction  of  the  pull  of  the 
string  exerted  upon  the  body. 

2.  In  the  preceding  example,  how  great 
a  horizontal  force  would  be  required  to 
deflect  the  cord  30°,  and  how  great  would 
the  corresponding  tension  be? 

3.  The  two  ends  of  a  cord  are  fastened  to 
hooks  in  the  same  horizontal  line,  and  at 
the  middle  a  second  cord  is  knotted  which  sustains  a  freely 
hanging  body  weighing  100  lbs.  The  distance  between  the 
hooks  being  such  that  the  first  cord  makes  angles  of  20°  with 
the  horizontal,  determine  the  tension  in  each  half  of  the  first 
cord.  Ans.  146  lbs. 

4.  Call  the  angle  in  the  preceding  example  a  and  the  weight 
of  the  body  W.     Deduce  an  expression  for  the  tension. 


6" 

p, 

'•  >o.'"|/^ 

/ 

§  II  ]  FLEXIBLE  CORDS.  105 

5.  Suppose  that  the  second  cord  in  ex.  3  is  knotted  to  the 
first  at  such  a  point  that  the  incHnations  are  10  and  50  degrees. 
Determine  the  tensions,  employing  the  special  condition  of 
equilibrium  of  art.  118.  Ans.  74.2  lbs.      I 

6.  Fig.  71  represents  an  upright  frame  within  which  is 
arranged  a  network  of  cords;  the  connec- 
tion EC  consists  of  a  spring  balance. 
Suppose  that  the  balance,  wrong  end  up, 
reads  10  lbs.  Determine  the  tension  in 
AF.  Ans.  26§  lbs. 

D  C 

7.  A  given  bar  is  to   be   supported  in  yig.  71. 
a  certain  position  by  two  strings  fastened 

to  its  ends.     What  condition  must  the  directions  of  the  strings 
satisfy?     (See  prop.,  art.  125.) 

8.  A  uniform  bar  weighing  no  lbs.  is  supported  in  a  hori- 
zontal position  by  strings  fastened  to  its  ends.  If  one  is  in- 
clined 30°  to  the  bar,  what  is  the  inclination  of  the  other  and 
what  are  the  tensions? 

9.  A  uniform  plate  triangular  in  shape  is  suspended  in  a 
horizontal  position  by  vertical  cords,  one  being  fastened  to 
each  comer.  Show  that  each  cord  sustains  one  third  the 
weight  of  the  plate. 

10.  Three  hooks.  A,  B,  and  C,  in  a  ceiling  lie  on  a  circle  of 
3  ft.  radius,  and  the  arcs  AB,  BC,  and  CA  are  equal.  A  small 
ring  is  suspended  at  a  point  4  ft.  below  the  centre  of  the  circle 
by  means  of  three  cords  fastened  to  the  hooks.  If  a  body 
weighing  100  lbs.  is  suspended  from  the  ring,  how  great  is  the 
tension  in  each  cord?  Ans.  4if  lbs.      t 

11.  Suppose  that  in  the  preceding  example,  the  arc  AB 
is  a  quadrant  and  AC  and  BC  are  equal.  Determine  the 
tensions. 

12.  It  is  required  to  determine  the  weights  necessary  to 
hold  a  cord  in  the  position  shown  in  fig.  72(a). 

Solution:  The  lines  radiating  from  O  (fig.  726)  are  drawn 
parallel  to  the  segments  of  the  cord,  and  AE  is  any  vertical 
line.  If  the  weights  W^,  W^,  W^,  and  W^  are  proportional  to 
AB,  BC,  CD,  and  DE  respectively,  they  will  hold  the  cord 
in  the  given  position.      For,  imagine  all  the  knots  except  the 


(b) 


1 06    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM,   [Ch! 

first  connected  by  vertical  cords  to  a  fixed  support  below,  and 
A^  a  body  kung  from  the  first  knot ;  the 
cord  is  then  in  the  required  posi- 
tion. According  to  some  scale,  AB 
cK^o  represents  W^,  and  according  to  the 
same  scale  BC,  CD,  and  DE  repre- 
sent the  tensions  in  the  imaginary 
vertical  cords  at  the  second,  third, 
and  fourth  knots  respectively.  But 
these  tensions  can  be  supplied  by 
bodies  suspended  from  the  knots. 

130.  Position  Assumed  by  a  Cord 
Sustaining  Loads. — Let  fig.  73  repre- 
sent the  nth  knot  on  the  cord,  the  load  suspended  from  which 
call  Wn  and  the  tensions  in  the  cord  segments  on  the  left 
and  right  Tn  and  T^+i  respectively.  Since 
the  three  forces  represented  are  in  equi- 
librium, ~^- 
„+i  CO  a^_^i- Tn  cos  an  =  o;  t^ 
sin  a^_^^  — Tn  sin  an  — Wn  =  o. 
From  the  first  equation,  it  follows  that  the 

horizontal  components  of  the  tensions  in  the  cord  segments  are 
equal;  that  is,  if  H  denotes  the  horizontal  component  in  any 
segment, 

if  =r„cos  «„==  r„+i  cos  a„+i  =  etc.     .     .     .     (i) 

This  equation  combined  with  the  second  one  above  gives 

ta.n  an+i=' tan  an -{-W„/H (2) 

By  means  of  this  equation  the  direction  of  each  segment  may 
be  computed  if,  in  addition  to  all  the  weights,  the  tension  in 
and  inclination  of  one  segment  or  the  inclinations  of  two  ad- 
jacent segments  are  known. 

EXAMPLES. 


IF,=  T 


I.  Three  bodies  weighing  100,  120,  and  200  lbs.  are  sus-  ; 
pended  in  the  order  named  from  three  knots  in  a  cord  which  is  I 
so  supported  that  the  second    segment  is  horizontal  and  its  I 


II.] 


FLEXIBLE  CORDS. 


107 


tension  is  160  lbs.     Determine  the   directions  of  the  other  seg- 
ments. Ans.  0:1  =  32°. 

2.  Suppose  that  the  supports  in  the  preceding  example  are 
such  that  the  middle  knot  is  the  lowest  and  that  the  segments 
adjacent  to  that  knot  are  inclined  30°  above  the  horizontal. 
Determine  the  inclinations  of  the  other  segments  and  all  the 
tensions. 

3.  Solve  the  preceding  examples  graphically. 

131.  The  Loads  are  Equal  and  Uniformly  Spaced  Horizontally. 
— The  knots  are  on  a  parabola;  proof  follows.  Let  0,  i,  2  .  .  .  w 
(fig.  74)  be  knots  on  the  cord  from 
which  the  loads,  W ,  are  suspended, 
and  let  x  and  y  axes  be  taken  as  in 
the  figure.  Denote  the  angles  which 
Oi,  12,  23,  etc.,  make  with  the  axes 
hy  a^,  a2,  a^  .  .  .  etc.,  the  horizontal 
distance  between  consecutive  loads 
by  s,  and  the  coordinates  of  the  wth 
knot  by  x  and  y.     Then 

x  =  ns\ (i) 

y  =  s  (tan  cKj+tan  a^-}- .  \  .). 

From  eq.  (2),  art.  130, 

tan  a^  =  tan  a^  +  W/H ; 


Fig,  74. 


Similarly, 


lience 


tan  0:3  =  tan  a^  +  W/H  =  tan  a^  +  2W/H. 


tan  a4  =  tan  a^  +  ^W/H,  etc.; 


y  =  sJntana^  +  in(n-i)W/H] (2) 

Combining  (i)  and  (2)  and  eliminating  n,  we  get 

2Hs 


2  tan  a^H  —  W 

W 


sx 


W 


(3) 


which  is  the  equation  of  a  parabola. 

If  the  loads  are  so  closely  spaced  as  to  constitute  a  practically 
continuous  load,  the  cord  is  practically  parabolic.  The  equa- 
;tion   of  the  parabola  referred  to  vertical  and  horizontal  axes 


io8    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

though  the  lowest  point  on  it  may  be  derived  from  eq.  (3),  or 

independently  as  follows:  ♦ 

Let  0  (fig.  75)  be  the  lowest  point  on  the  cord,  Q  any  other 
point,  and  let  w  denote  the  load  per 
unit  horizontal  distance ;  then  OQ 
sustains  a  load  wx.  The  tension  at 
0  call  H,  and  that  at  Q,  T.  The 
three  forces  applied  to  OQ  are  con- 
current, and  their  equilibrium  equa- 
tions are 


wx 
Fig. 


75- 


IF^  = 


■H  +  Tcosd  =  o; (i) 

■wx-\-T  sin  6  =  0 (2) 

Combining  these  with  tan  d  =  y-^x/2,  we  get 


^Fy  = 


2     2f/ 

^  =  — y, 

w 


(3) 


which  is  the  equation  sought. 

If  the  points   of  suspension  are  at  the  same  level,  then   (see 
fig.  76  and  eq.  (3)), 

d  =  way2H (4) 


EXAMPLES. 

1.  A  cord  is  supported  at  two  points  on  the  same  level  30 
ft.  apart,  and  its  lowest  point  is  8  ft.  below  the  level  of  the 
supports.  If  the  load  is  20  lbs.  per  ft.,  what  are  the  tensions 
at  the  supports  and  at  the  lowest  point  ?         Ans.  if  =  281 J  lbs. 

2.  Suppose  that  one  support  is  3  ft.  higher  than  the  other 
and  that  the  lowest  point  of  the  cord  is  8  ft.  lower  than  the 
lower  support.     What  is  the  greatest  tension  in  the  cord? 

132.  Position  Assumed  by  a  Heavy  Flexible  Cord  Suspended 
from  Two  Points. — The  curve  assumed  by  such  a  cord  if  uni- 
form in  weight  (and  such  only  are  discussed  below)  is  called  a. 


§n.] 


FLEXIBLE  CORDS. 


109 


common  catenary.     Let  AB  (fig.  77)  represent  such  a  cord,  C 
its  lowest  point,  Q  any  other  point,  5  the  length  CQ ,  H  and  T 


Fig.  77. 
the  tensions  at  C  and  Q  respectively,  and  w  the  weight  of  the 
cord  per  unit  length.     Then  the  forces  on  the  portion  CQ  are 
if,   r,  and  ws  acting  as  shown  in  fig.   77(a);    hence 

JFx=  —H  +  T  cos  <j)  =  o; 

IFy  =  —  W5  +  r  sin  ^  =  o. 
Prom  these,  by  division,  we  get 

tan  (f)  =  ws/H. 
Por  convenience,  let  c  denote  a  length  of  the  cord  whose  weight 
equals  the  tension  at  C,  then  H/w  =  c\  and  since  tan  (j>  =  dy/dx, 

%-!■ (^) 

Integration  of  equation  (i)  leads   to  the   equation   of  the 
.catenary.     Differentiating,  we  get 

.<S  =  v-  

Since  {dsy  =  (dyy+(,dx)\     ds  =  dx'^i+{dy/dxy\ 

Integrating,  we  get 

Now  dy/dx  =  o,  where  x  =  o  (see  fig.),  hence  C'  =  o. 


hence 


or 


no    /iPPUCATlONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 
Solving  the  equation  for  dy/dx,  we  find  that 


dy 
d^ 


=  i(W^-^-V^), (2) 


e  being  the  base  of  the  Naperian  system  of  logarithms. 
Integrating  the  last  equation,  we  get 


j  =  _(^^/c+^-^A)  _j_(7//^ 


Since  y  =  c  when  x  =  o  (see  fig.),  the  constant  C"  is  zero;  hencejj 
the  equation  of  the  catenary  referred  to  the  axes  of  the  figure  is  \ 


y  =  —  (^e^/c-^e~^/^). 


(3) 


Several  interesting  relations  may  be  deduced  as   follows :  ] 

From  (i)  and  (2)  i 

c  ^ 

s  =  -(e''/c-e-x/c) (4)1 

2  i 

Combining  (3)  and  (4),  we  find  that  J 


Also, 


y+s  =  ce''/'^,     or 


^log^ 


y+s 


(5) 
(6) 


From  the  equilibrium  equations,  i 

72  =  ^2(^2  +  ^2);  i 

hence  T  =  wy (7)  ■ 

EXAMPLES.  ] 

I.  A  measuring-tape  400  ft.  long  weighing  0.005  lt)s.  per  ft.  ] 
y  is    suspended  from  two  points,  A  and  i 

By  the  supporting  pulls  there  being  re-  \ 
spectively  1.6  and  2.0  lbs.  Compute  I 
the  horizontal  and  vertical  distances  \ 
between  A  and  B. 

Solution :  Let  T^  and  7^  denote  the  ^ 

pulls    at   A  and    B,  and    /^   and  /j  the  \ 

X    lengths  AC  and  BC  respectively  (fig.  78).  \ 

Fig.  78.  From  eq.  (7),  I 

Tr  =  wh^y    or     61  =  1.6^0.005  =  320  ft.,  \ 


§11.]  FLEXIBLE  CORDS.  HI 

and  T2  =  wb2,     or     62  =  2.0 -=-0,005  =400  ft.; 

hence  62  —  61  =  80  ft. 

From  eq.  (6),  a^  =  cloge— — -, 

and  -'a^=c  \oge  — — -- 

c 

To  determine  a^  and  flj  values  of  /j,  /j,  and  c  are  needed. 

Fromeq.  (5),  b{^  =  li^  +  c^=  102  400 

and  62^  =  ^2^  +  ^^  =  1600005 

also,  /i -I- /a  =  400. 

Solution  of  these  three  equations  shows  that 

/i=i28,     /2  =  272,     and     ^  =  293+ ft., 
which  values  if  substituted  in  the  expressions  for  a^  and  flj  give 

ai=i22.7     and    03  =  242.7  ft.; 
hence  ai+a,  =  365.4  ft. 

2.  What  is  the  tension  at  the  lowest  point  of  the  tape  in 
ex.  I  ? 

3.  A  tape  whose  length  is  100  ft.  weighs  0.005  lbs.  per  ft. 
and  is  subjected  to  end  pulls  of  10  lbs.,  the  ends  being  at  the 
same  level.  Compute  the  distance  between  the  ends  and  the 
"sag."  Ans.  Distance  =  99.996  ft. 

Sag  =  0.625  ft. 

4.  A  tape  is  supported  at  two  points  on  the  same  level,  its 
length  =  2/,  its  weight  per  unit  length=w  and  the  end  pulls  =  P. 
If  2a  and  d  denote  distance  between  supports  and  the  sag  re- 
spectively, show  that 

d=p/w-(pyw'-i')K 


112   APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.     [Ch.  VI J 

5.  Suppose  that  a  chain  200  ft.  long  is  suspended  fromj 
two  points  on  the  same  level  and  59  ft.  apart.     Find  the  sag.        \ 

Solution:  This  example  can  be  solved  by  trial  only;  thus,i 
referring  to  fig.  77,  it  is  seen  that  for  B,  ^  =  25  and  5  =  100  ft.;; 
hence  eq.  (4)  becomes 

=J(^25/(r_^-2S/c)^  I 

i 

It  will  be  found  that  c  =  7 .65  will  nearly  satisfy  this  equation.* ; 
Then  eq.  (5)  for  B  becomes  \ 

:V^  =  100^  +  72.65,     or     >'  =  ioo.28;  \ 

hence  the  sag  is  \ 

100.28  —  7.65,     or     92.63  ft. 

i 

6.  Determine  the  length  of  a  chain  which  sags  20  ft.  if] 
suspended  from  two  points  on  the  same  level  and  80  ft.  apart.      \ 

Solution:  This  example  can  be  solved  by  trial  only;  thus,' 
referring  to  fig.  77,  it  is  seen  that  for  By  x=^o  and  y=c +  2o,\ 
hence  eq.  (3)  becomes 

c  +  20  = — (^40/c  4- ^  -  40A) , 

i 

It  will  be  found  that  ^=42.7    nearly  satisfies  this  equation;; 
hence  eq.  (5)  for  B  becomes  ; 

(42.7  +  20)2=52  +  422.7,  ] 

or  5=45.9.     The  length  of  the  chain  is  therefore  91.8  ft.  j 

133.  Approximate  Equations, — If  the  points  of  suspension  1 
are  at  the  same  level  and  the  sag  is  small,  the  length  of  the  cord  ] 
and  distance  between  supports  are  nearly  equal,  and  the  num-  ] 
bered  equations  below  are  close  approximations.  They  may  i 
be  deduced  as  follows:  \ 

Since  the  slope  of  the  cord  is  everywhere  small,  the  load  1 
per  horizontal  unit  length  is  nearly  uniform;  hence  the  catenary  ; 

*  The  determination  of  c  in  exs.  5  and  6  is  much  facilitated  by  use  \ 
of  a  table  of  hyperbolic  functions,  which  gives  values  of  \{ex/c—e-x/c)^  or  : 
sinh  x/c,  and  ^{ex/c-\-e-*/c),  or  cosh  x/c,  for  different  values  of  x/c.  ; 

i 


§111.]  TACKLE.  113 

is  very  nearly  parabolic  and  the  equations  deduced  in  art.  131 
must  be  nearly  correct  for  a  flat  catenary. 
From  eq.  (4)  of  that  article, 

d=wa?l2R (i) 

In  works  on  calculus  it  is  shown  that  the  length  of  a  parabolic  arc, 
as  OB  (fig.  76),  is  given  by 

H 


1  = 

w 


tan  aVi  +tan2a+log^(tan  a:  + Vi  ^-tan^a)    , 


or,  developed, 


,     H/  ,  tan^a  \ 

/  =  —   tan  q:H f-.  .  .    . 

w\  6  / 


Since,  in  a  parabola,  tan  a  =wa/H,  approximately, 

l=a{i+w^a^/()P^), (2) 

and  a=l{i-wH^/(>P^) (3) 

EXAMPLES. 

1.  Solve  ex.  3,  art.  132,  by  the  approximate  equations  of 
this  article.  Ans.  (i)  gives  for  sag  0.625  ^t. 

(3)      "      "  distance  99.99    ** 

2.  Hard  copper  wire  weighs  3.85 A  lbs.  per  foot,  A  being 
the  area  of  its  cross-section  in  square  inches.  If  such  a  wire 
be  suspended  between  two  points  at  the  same  level  200  ft.  apart, 
and  the  maximum  pull  on  it  be  limited  to  20000  lbs.  per  sq.  in., 
what  is  the  proper  length  of  the  wire  and  the  sag? 

Ans.  Sag  =  0.96  ft. 

3.  The  rope  of  a  rope  drive  when  not  running  is  observed 
to  sag  I  ft.,  and  the  distance  between  the  centres  of  the  wheels 
is   70  ft.     What  is  the  maximum  tension  in  the  rope? 

§  III.     Tackle. 

134.  The  Pulley. — A  wheel  with  a  flat-faced  or  grooved  rim 

which  can  rotate  about  its  axis  is  often  used  as  a  "power 
transmission"  device;  when  so  used  it  is  called  a  pulley.  It 
may  rotate  freely  about  an  axle  or  shaft,  and  is  then  called  a 


o 


114    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI.  1 

loose  pulley;  or  it  may  be  fastened  to  a  shaft  which  turns  in  I 
bearings  with  the  pulley,  when  it  i&  called  a  fast  pulley. 

A  combination  of  one  or  more  grooved  pulleys  and  a  rope  ; 
or  chain  used  for  raising  heavy  bodies,  is  called  a  tackle.  The  I 
pulley  or  pul  eys  rotating  on  the  same  axle  together  with  the  I 
frame  supporting  the  axle  is  called  a  block.  A  pulley  is  called  ■ 
movable  or  fixed  according  as  the  block  of  which  it  is  a  part  1 
does  or  does  not  move  while  the  load  is  raised  or  lowered.  \ 

135-  Tension  in  a  Cord  on  a  Pulley. — Fig.  80  represents  a  ; 
single  pulley,  loose  or  fast,  and  fixed  or  mov-  : 
able,  about  a  part  of  whose  rim  a  flexible  cord  ; 
\^  or  belt  is  tightly  wrapped.  We  wish  to  find  \ 
"■""  the  relation  between  the  tensions  in  the  cord  or  ; 

Fig.  80.  "belt  on  opposite  sides  of  the  pulley  on  the  sup-  \ 
position  that  the  rubbing  surfaces  at  the  axle  are  ' '  smooth. ' '       i 

If  the  rubbing  surfaces  at  the  axle  are  smooth,  then  the  ' 
pressure,  P,  of  bearings  on  the  axle  (fast  pulley)  or  the  pressure  \ 
of  the  axle  on  the  pulley  (loose)  is  such  that  its  line  of  action  \ 
passes  through  the  axis  of  the  pulley  (see  art.  138).  Then  of  ' 
the  three  forces  7',  T",  and  P,  the  only  ones  having  a  moment  i 
about  the  axis  are  T'  and  7",  and  if  the  cord  or  belt  is  flexible  | 
their  arms  are  equal;  hence  T'  =  T".  That  is,  the  tensions  in  \ 
a  cord  or  belt  on  opposite  sides  of  a  pulley  which  the  cord  or  belt  \ 
encircles  are  equal  if  the  cord  or  belt  is  flexible  and  the  rubbing  \ 
surfaces  at  the  axle  are  smooth.  ^ 

In  the  following  examples  this  relation  is  made  use  of,  the  ; 
cords  being  assumed  flexible  and  the  rubbing  surfaces  smooth.  ; 
This  assumption  is  far  from  the  truth  in  actual  tackle,  and  it  | 
should  be  remembered  that  the  results  obtained  below  are  for  ; 
ideal  cases.  Axle  friction  and  rigidity  of  cordage  are  discussed  i 
later.  ; 

EXAMPLES.  ] 

1.  In  the  arrangements  shown  in 'fig.  81  (a)  and  (6),  regard  ■ 
F  and  a  as  given  and  compute  W  and  0.  \ 

2.  Compute  the  pressure  upon  the  axle  (fig.  82),  a  and  W  \ 
being  given.  ! 

3.  What  is  the  relation  between  F  and  W  in  the  tackle  I 


§  "I.] 


TACKLE. 


115 


shown  in  fig.  83,  neglecting  the  weight  of  the  parts?     (Consider 

the  equiHbrium  of  the  part  below  the  dotted  W 

line.) 

4.  What  is  the  relation  between  F  and  W 
in  the  tackles  shown  in  figs.  84  and  85,  the 


(a)  (b) 

Fig.  81. 


Fig.  87.      Fig.  86. 


weight  of  the  parts  being  neglected?     What  is  the  tension  in 
the  cord  supporting  the  tackle  of  fig.  85? 

Ans.  For  fig.  85,  W  =  ()F. 

5.  What  is  the  relation  between  F  and  W  in  the  tackle 
shown  in  fig.  86?     What  are  the  pulls  on  the  hooks  at  A  and  jB? 

6.  Fig.  87  represents,  in  principle,  a  Weston  differential 
tackle.  The  wheels  in  the  upper  block  are  fastened  together, 
are  of  unequal  diameters,  and  their  grooves  are  made  so  that 
the  hoisting  chain  will  not  slip  in  them.  Determine  the  rela- 
tion between  F  and  W,  and  the  pull  at  the  upper  hook. 

Solution:  Consider  the  equilibrium  of  the  part  of  the  tackle 
above  the  dotted  line.     The  external  forces  upon  it  are  four  in 


Ii6     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

number  if  its  weight  be  neglected,  namely,  the  upward  pull  at 
the  hook,  the  pull  F,  the  tension  iif  the  chain  on  the  left  of  the 
smaller  wheel  (W/2),  and  the  tension  in  the  chain  to  the  right 
of  the  larger  wheel  (W/2).  One  condition  of  equilibrium  for 
these  forces  is  that  their  moment  sum  with  respect  to  the  axis 
of  the  wheels  is  zero;  hence  if  r'  and  r"  denote  radii  of  smaller 
and  larger  pulleys, 

Fr"  + 1^72-1^/72-0, 
or  W=' 


r"  -r 


7.  Suppose  that  in  fig.  81(6),  a  is  zero  and  that  a  man  whose 
weight  is  W  sits  upon  the  load  and  exerts  a  pull  F.  How 
great  must  the  pull  be  to  raise  man  and  load?  Is  it  greater  or 
less  than  Wl 

§  IV.     Smooth  Supports. 

136.  Definitions. — It  is  known  from  experience  that  to  slide 
one  body  over  another  even  at  constant  speed  requires  the  appli- 
cation of  more  or  less -force;  also  that  a  moderate  "sliding 
force ' '  may  not  cause  a  body  to  move.  It  is  inferred  that  the 
second  body  exerts  a  force  upon  the  first  which  is  opposed  to 
the  sliding. 

More  definitely,  let  fig.  88  represent  two  bodies  whose  sur- 
face of  contact  is  a  horizontal  plane,  the  upper 
one  being  subjected  to  a  horizontal  force  P.  The 
;r  lower  body  exerts  upon  the  upper  a  force  such 
Fig.  88.  ^^  -^'  "^^^  horizontal  component  of  which,  R^,  is 
the  resistance. to  sliding.  If  the  upper  body  is  in 
equilibrium,  Rx  =  P.  Now  it  is  known  that  the  smoother  the 
surfaces  of  contact,  the  smaller  is  the  force  required  to  cause 
sliding  and  hence  the  smaller  the  resistance  to  sliding.  We  are 
thus  led  to  the  conception  of  a 

Perfectly  smooth  surface,  which  may  be  defined  as  one  which 
offers  no  resistance  to  the  sliding  of  a  body  upon  it.  Such  a 
surface  is  ideal,  but  there  are  surfaces  which  are  nearly  perfectly 
smooth. 


§  IV.]  SMOOTH  SUPPORTS.  117 

For  simplicity,  we  will  assume  the  surfaces  of  contact  in 
some  of  the  following  examples  as  perfectly  smooth.  For 
brevity,  they  will  be  called  smooth  surfaces,  while  those  whose 
resistance  to  sliding  is  to  be  taken  into  account  will  be  called 
rough.* 

137.  Reaction  of  a  Smooth  Surface. — A  perfectly  smooth 
plane  surface  can  exert  a  force  only  along  a  normal  to  it.  For, 
by  definition,  such  a  surface  offers  no  resistance  to  the  sliding 
of  a  body  over  it,  that  is,  the  reaction  which  the  surface  offers 
has  no  component  along  the  surface  and  hence  that  reaction 
must  be  directed  along  the  normal.  If  the  surface  is  not  a  plane 
one,  then  the  resistance  exerted  by  each  elementary  part  of 
the  surface  is  directed  along  the  normal  to  that  part. 

EXAMPLES. 

I.  A  body  weighing  100  lbs.  rests  upon  a  smooth  plane 
inclined  at  an  angle  of  30°  with  the  horizon,  and  is  prevented 
from  slipping  by  a  cord  fastened  to  it  which  leads  off  up  the 
incline  and  over  a  smooth  pulley  at  the  top  and  supports  a 
body  which  hangs  freely  from  that  end.  Determine  the  weight 
of  the  suspended  body  and  the  resistance  of  the  plane. 

Solutions:  Consider  the  forces  applied  to  the  body  upon 
the  inclined  plane ;  they  are  three  in  number,  namely,  its  weight, 
the  pull  of  the  cord,  and  the  resistance  of  the  incline.  The 
system  is  coplanar  and  concurrent;  and  the  pulley  being  smooth, 
the  pull  of  the  string  equals  the  weight  of  the  suspended  body. 

(i)  Algebraical.  Let  W  denote  the  weight  of  the  sus- 
pended body  and  R  the  resistance  of  the  plane. 
The  forces  acting  upon  the  body  are  represented 
in  fig.  89.  For  such  a  system  there  are  two 
equilibrium  equations  (art.  125),  and  if  the  x 
^and  y  axes   be  taken  horizontal  and  vertical  re-    iooit>s. 


spectively, 


Fig.  89. 


IF  J,  =  VP^  cos  30°  -i?  cos  60°  =  o ; 
IFy  =  W  sin  7^0° -\-R  sin  60°  — 100  =  0. 


*  Rough  surfaces  are  discussed  in  §  VIII. 


Ii8     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

These  equations  determine  W  and  R.  (The  student  should 
select  the  coordinate  axes  parallel  *and  normal  to  the  incline, 
write  the  equilibrium  equations  and  compare  the  length  of 
their  solution  with  that  of  those  above.) 

(2)  Graphical.     The  condition  of   equilibrium   is   that   the 

polygon  for  the  three  forces  must 
close  (art.  125).  To  construct  the 
polygon,  the  wholly  known  force 
(the  weight  of  the  body)  is  repre- 
sented first;  thus,  AB  (fig.  90)  is 
drawn  vertically  and  one  inch  long. 
Next,  a  line  from  A  (or  B)   parallel 

^  to  the  incline,  and   one  from  B  (or 

Fig.  90. 

A)   parallel    to   the  normal  to  the 

plane  are  drawn,  thus  determining  C.  Then  BC  and  CA  rep- 
resent the  magnitudes  and  directions  of  the  forces  sought. 

2.  Solve  ex.  (i),  supposing  that  the  pulley  is  so  placed  that 
the  cord  leads  off  horizontally  from  the  body  upon  the  inchned 
plane.  Ans.  1^  =  57  .  7  lbs. 

3.  Solve  ex.  (i),  supposing  that  the  pulley  is  so  placed  that 
the  cord  leads  off  from  the  body  upon  the  inclined  plane  at  an 
angle  of  50°  with  a  horizontal. 

4.  Two  planes  which  are  inclined  at  angles  a  and  /?  with  a 
horizontal  plane  intersect  in  a  horizontal  line,  and  a  cylinder 
whose  weight  is  W  rests  between  and  upon  them.  Compute 
the  pressure  on  each  plane. 

5.  A  bar  24  in.  long  rests  in  a  smooth  hemispherical  bowl 
30  in.  in  diameter.  The  centre  of  gravity  of  the  bar  is  10  in. 
from  the  lower  end.     Determine  the  position  of  equilibrium. 

Ans.  Inclination  of  bar  to  horizontal  is  12°  32', 
[Suggestion:  The  three  forces  maintaining  the  equilibrium  of 
the  bar  are  coplanar  and  concurrent.  The  action  lines  of  two, 
the  pressures  of  the  bowl  on  the  bar,  intersect  at  the  centre  of 
the  bowl  and  therefore  the  centre  of  gravity  of  the  bar  must  be 
vertically  below  the  centre  of  the  bowl.  Make  a  sketch  show- 
ing this  relation,  mark  the  known  lengths,  and  solve  trigonomet- 
rically  for  the  inclination  of  the  bar.] 

6.  A  uniform  bar  whose  length  is  40  in.  is  supported  by  a 


§  IV.]  SMOO TH  SUPPOR TS.  119 

smooth  vertical  wall  and  a  smooth  peg  whose  axis  is  horizontal, 
parallel  to  the  wall,  and  15  in.  therefrom.  Determine  the  posi-, 
tion  of  equilibrium.  Ans.  Angle  with  vertical  is  65°  18'. 

138.  Pin  Joint  or  Hinge. — In  some  of  the  following  exam- 
ples reference  is  made  to  bodies  joined  by  means  of  a  pin  hinge. 
Such  a  joint  may  consist  of  a  cylindrical  pin  and  two  cylindri- 
cal holes,  one  in  each  of  the  bodies  joined,  into  which  the  pin  is 
inserted  allowing  relative  rotation  about  the  pin.  Still  more 
simply,  the  joint  may  consist  of  a  pin  which  is  rigidly  fastened 
to  one  body  and  inserted  into  a  hole  in  the  other,  allowing 
rotation. 

If  the  cylindrical  surfaces  of  the  pin  and  hole  are  smooth, 
the  forces  exerted  upon  these  surfaces  act  normally,  and  their 
resultant,  therefore,  passes  through  the  axis  of  the  pin. 

EXAMPLES. 

1.  The  body  ABC  (fig.  91)  is  supported  by  a  smooth  hinge 
at  A   and   a   smooth   surface   at  B,  and 
a    horizontal    force  F  is   applied   at   C. 
Compute  the  reactions  of  the  supports, 
neglecting  the  weight  of  the  body. 

Ans.  Reaction  at  5  is  Fa//. ,  "^^^^  ^^* 

2.  Suppose  that  the  support  of  the  bar  at  B  in  ex.  i  is  also 
a  smooth  pin  hinge  in  the  same  horizontal  plane  with  the  other 
one.     Determine  the  supporting  forces. 

Solution:  The  reactions  at  M  and  N  and  the  force  F  to  be 
in  equilibrium  can  not  be  parallel,  hence  they  must  be  con- 
current (art.  125);  also  their 
force  polygon  must  close.  Any 
reactions  fulfilling  these  two  con- 
ditions will  balance  F.  Many 
pairs  of  reactions  can  satisfy  the 
F         B  conditions — for  example,  BC  and 

^^^-  92.  CA,  or   BC    and   C'A  (fig.  92); 

hence  the  problem  is  not  statically  determinate. 

The  vertical  components  of  the  reactions  are  determinate, 
which  may  readily  be  proved  thus:    Call  the  reactions  at  M 


if 

-^->nc 

5 

a 

A 

'                 ..i 

"^c'li^ 

- 

--r"^' 

(^■—-^ 

~^ 

M 

N 

7f 

I20    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

and  A^,  R'  and  i?"  respectively,   and  imagine  them  resolved 
horizontally  and  vertically.     The  system  F,  RJ ,  Ry  ,  Rx\  Ry^ 
is  coplanar    non- concurrent    non-parallel,  and  there  are  three 
algebraic  conditions  of  equilibrium  (art.  125).     They  are 
IFx^F+RJ+RJ'  =  o; 

IFy  =  Ry'+Ry"=0', 

IMM  =  Ry'l-Fa  =  o. 
From  the  last  two  equations  it  follows  that 

Ry''=^Fa/l     and     Ry' = -Fa/l 
The  indeterminateness,  therefore,  is  really  with  the  horizontal 
components;  their  arithmetic  sum  or  difference  equals  F,  as 
may  be  seen  from  the  first  equation  or  from  the  force  polygon 
in  fig.  92. 

3.  A  uniform  bar  weighing  200  lbs.  and  10  ft.  long  leans 
against  a  smooth  vertical  wall,  and  its  lower  end  is  fastened 
to  a  floor  by  a  smooth  pin  hinge  whose  axis  is  horizontal  and 
parallel  to  the  wall.  The  distance  of  the  hinge  from  the  wall 
being  8  ft.,  determine  the  forces  supporting  the  bar,  solving 
graphically. 

Ans.  Force  on  upper  end  is  horizontal  and  equals  133  J  lbs. 

§  V.  Three  Typical   Problems    on  Coplanar  Non-concur- 
rent Forces. 

139.  Problem  I. — The  forces  of  a  parallel  system  in  equi- 
librium are  completely  known  except  two  whose  action  lines  only 
are  known.     It  is  required  to  determine  completely  these  two  forces. 

Algebraic  Solution. — There  are  two  conditions  of  equilibrium 
for  this  system  (art.  125),  namely, 

IF    =0     and     IM   =0, 
or  IMa=o     and     IMi,=o.* 

Either  set  of  equations  furnishes  the  solution.     It  is  advan- 
tageous to  select  moment  origins  on  the  lines  of  action  of  the 

*  Whenever  a  force  whose  sense  is  unknown  is  to  be  entered  into  a 
resolution  or  moment  equation,  a  sense  should  be  assumed  for  that  force 
and  adhered  to  in  the  solution  of  the  equation.  The  correct  sense  is 
indicated  by  the  sign  of  the  computed  value  of  that  force.  It  is  as 
assumed  or  opposite  according  as  the  sign  is  positive  or  negative. 


§v.] 


THREE   TYPICAL  PROBLEMS. 


12? 


unknown  forces.  Then  each  moment  equation  will  contain 
but  one  unknown  quantity  and  may  be  readily  solved. 

Illustration:  Let  it  be  required  to  determine  the  magni- 
tudes and  directions  of  the  two  forces 
F'  and  F"  (fig.  93),  all  the  forces  there 
represented  being  in  equilibrium,  F^, 
F2,  and  F3  being  700,  300,  and  500  lbs., 
and  a,  b,  c,  and  d  i,  3,  5,  and  2  ft. 
respectively. 

If  F'  and  F''  are  assumed  to  act  upward  and  the  moment 
origin  is  taken  on  F", 

2'F=  —  700  +  F'  — 3oo  +  5oo  +  F"  =  o; 
JM=  +700-  II  —F'- 10  +  300-7  —  500-2  =0. 

From  the  second  equation,  F'  =  88o  lbs.,  and  this  value  sub- 
stituted in  the  first  gives  F"=— 380  lbs.  The  minus  sign 
means  that  F"  does  not  act  up  as  assumed,  but  down. 

Employing  the  second  set  of  equilibrium  equations,  and 
selecting  moment  origins  on  F'  and  F"  respectively, 

IM  =  joo-i    — 3oo-3  +  5oo-8  +  F"-io  =  o; 
2'M=  700- II —F'- 104-300- 7— 500-2=0. 

These  solved  give  the  same  values  as  those  found  above. 

Graphical  Solution. — The  conditions  of  equilibrium  are  that 
the  force  and  funicular  polygons  close  (art.  125).  In  the 
process  of  constructing  them  the  unknown  quantities  will  be 
determined. 

Illustration:  Let  the  data  be  the  same  as  that  in  the  pre- 
ceding illustration.  First  the  force  polygon  should  be  drawn 
as  far  as  possible;  thus,  AB,  BC,  and  CD  (fig.  94)  representing 
Fj,  F2,  and  F3  respectively.  If  either  one  of  the  unknown 
forces  be  lettered  DE,  the  other  must  be  EA  because  the  poly- 
gon must  close.  It  remains  to  locate  the  point  F;  this  is  done  by 
means  of  the  funicular  polygon.  Before  beginning  to  draw  the 
polygon,  recall  what  the  strings  represent  (see  arts.  37  and  38). 
If  the  polygon  be  begun  on  ab,  strings  oa  and  ob  must  be  drawn 
from  that  point ;  then  oc  is  drawn  from  where  ob  intersects  be, 
and  od  from  where   oc   intersects  cd.      Now  oa  is  the  action 


122     APPLICATIOh^S  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

line  of  one  of  the  components  of  EA,  and  should  be  extended 
to  ca,  thus  determining  a  point  /  in  ihe  action  line  of  the  other 
component,  EO.     Also,  od  is  the  action  line  of  one  of  the  com- 


Scale:l  ln.=  6  ft.  Scale.  1  in.=  100  lbs. 


(a) 


(b) 


Fig.  94. 


ponents  of  DE,  and  should  be  extended  to  de,  thus  deter- 
mining a  point  //  in  the  action  line  of  the  other  component, 
OE.  Now,  if  the  funicular  polygon  is  to  be  closed,  the  action 
lines  of  EO  and  OE  must  coincide  (see  art.  40) ;  hence  there  is 
but  one  string  oe,  and  it  passes  through  points  /  and  //.  The 
ray  corresponding  to  oe  may  now  be  drawn,  thus  determining  E. 


EXAMPLES. 

1.  Reverse  F^  and  F^  (fig.  93)  and  solve  algebraically , the 
example  used  for  an  illustration  above. 

2.  Solve  the  preceding  example  graphically. 

140.  Problem  II. — The  forces  of  a  non-parallel  system  in 
equilibrium  are  completely  known  except  two.  Of  these,  the  action 
line  of  one  and  a  point  in  that  of  the  other  are  known.  It  is  re- 
quired to  determine  completely  these  two. 

Algebraic   Solution. — Thdre    are    three    conditions    of   equi- 
librium for  a  system  of  this  kind  (art.  125),  namely, 
IF:,  =  o,       IFy  =  o,     IM   =0; 

or,  ^Ma  =  o,      IMb  =  o,      IMc  =  o. 

Either  set  furnishes  the  solution,  the  three  unknowns  being 
the  magnitude  (and  sense)  of  one  force,  the  magnitude  (and 
sense)  and  'nclination  of  the  other. 


§  v.]  THREE   TYPICAL  PROBLEMS.  123 

Illustration :  Suppose  a  horizontal  beam  which  is  supported 
at  one  end  by  a  pin  hinge  and  at  the  other  by  an  inclined  cord 
sustains  given  loads,  and  that  it  is  required  to  determine  the  ten- 
sion in  the  cord  and  the  pin  reaction. 

Let  fig.  05  represent  the  beam,  loading,  notation  etc.;   the 

ff /\Hinge D 

.-x-:^..:-..2x....^-i.>::^— -2^.— . 


TT^ 


50  lbs.  100  lbs.     25  lbs. 
Fig.  95. 

I  co-lb.  force  may  be   the  weight  of  the  beam.     The  unknowns 
are  denoted  by  Q,  P,  and  <9.*     Employing  the  first  set  of  equi- 
;  librium  equations  and  selecting  the  x  and  y  axes  horizontal  and 
vertical  respectively,  and  a  moment  origin  at  A, 

IF^=-P  cos  d  +  Q  cos  6o°  =  o; (i) 

i'Fj/  =  P  sin  ^-|-Q  sin  60°  — 50  — 100  — 25=0;     .     .     (2) 
2'Ma  =  Q- 6  sin  60°  — 50-2  — 100-3  — 25-5  =  0.    •     •     (3) 
These  equations  furnish  the  solution.     Observe  that  by  select- 
ing A  as  moment  origin,  two  unknowns,  P  and  6,  were  elimi- 
\  nated  from  the  moment  equation. 

Usually  it  is  advantageous  to  replace  the  force  whose 
magnitude  and  direction  are  unknown  by  two  rectangular 
components  acting  at  the  known  point  in  the  action  line  of 
'  the  force.  Then  the  unknowns,  instead  of  P  and  d  of  the  illus- 
vtration,  would  be  Px  and  Py,  and  when  these  latter  become 
[known,  P  and  6  are  easily  computed.  Such  substitution  trans- 
f  forms  the  problem  to  Problem  III  (art.  141).  The  example 
;  employed  above  as  illustration  is  solved  in  art.  141  on  this  plan. 
;  Graphical  Solutions. — (i)  The  conditions  of  equilibrium  are 
that  the  force  and  funicular  polygons  close  (art.  125).  In  the 
process  of  constructing  the  two  polygons  the  unknowns  will 
be  determined. 

*  In  this  example  it  is  evident  that  the  senses  of  P  and  Q  are  as  repre- 
sented and  that  ^<9o°.  When  the  senses  of  P  and  Q  and  "quadrant"  of 
6  are  not  evident,  they  should  be  sketched  as  they  appear  to  be,  and  then 
the  equilibrium  equations  should  be  written.  The  correct  senses  may 
be  inferred  as  explained  in  the  foot-note,  p.  120,  and  the  quadrant  of  6 
will  appear  from  the  solution. 


124    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 


0^- 


:- i._a_. 


h  t> 


v^ 


""^->.  ^ 


c  - 


Scale:Iin.=  4ft. 

(a)        .  A 


Scale:lin.=2001bs, 
Fig.  96. 


Illustration:  Let  the  data  be  the  same  as  in  the  preceding 

illustration.  First,  the  force  poly- 
gon is  drawn  as  far  as  possible; 
thus,  AB,  BC,  and  CD  (fig.  96), 
representing  the  50,  100,  and 
25  lb.  forces,  and  then  an  in- 
definite line  parallel  to  the  cord 
from  D  (or  A).  The  fourth  force, 
tension,  then  is  to  be  lettered 
DE,  and  the  remaining  one  must 
be  marked  EA,  since  the  poly- 
gon is  to  be  closed.  It  remains 
to  determine  E;  this  is  done  by 
means  of  the  funicular  polygon. 

If  the  polygon  be  begun  at 
any  point  of  ab,  be,  cd,  or  de, 
it  can  not  be  completed,  as  is  plain  from  an  examination 
of  the  unlettered  funicular  polygon  which  was  begun  from  a 
point  of  ab  taken  at  random.  The  difficulty  is  in  determining 
the  desired  intersection  of  the  string  corresponding  to  OA  and 
the  unknown  action  line  of  EA\  this  point  corresponds  to 
point  /  of  fig.  94.  If  the  funicular  polygon  be  begun  at  the 
given  point  of  the  unknown  action  line,  the  difficulty  is  avoided; 
thus,  oa  is  drawn  through  that  point,  and  then  ob,  oe,  and  od. 
Now  /  and  //  respectively  are  points  in  the  action  lines  of  the 
forces  EO  and  OE,  and  if  the  funicular  polygon  is  to  be  closed, 
the  lines  of  action  of  EO  and  OE  must  coincide  (art.  45);  hence 
there  is  but  one  string  oe,  and  it  passes  through  points  /  and 
//.  The  ray  corresponding  to  OE  may  now  be  drawn,  and  thus 
the  point  E  is  determined.. 

(2)  Find  the  resultant  of  the  wholly  known  forces,  and 
consider  the  system  as  consisting  of  that  force  and  the  two 
unknown  ones.  'Then,  making  use  of  the  fact  that  the  three 
forces  are  parallel  or  concurrent  (see  art.  125),  determine  the 
unknowns.* 


*  If  the  resultant  of  the  known  forces  is  a  couple,  the  two  unknowns  \ 
must  also  constitute  a  couple.  How  might  one  determine  graphically  1 
the  forces  of  the  latter  couple?  ^ 


§v.] 


THREE   TYPICAL  PROBLEMS. 


125 


Illustration:  Let  the  data  be  the  same  as  that  of  the  pre- 
ceding illustration.     The  resultant  of  AB,  BC,   and  CD  (fig. 


Fig.  97. 

97)  is  found  to  be  AD.  Extending  ad  and  .de  we  find 
their  intersection  which  is  one  point  in  the  action  line  of 
the  other  unknown  force;  another  point  in  its  action  line 
being  given,  the  centre  of  the  pin,  the  line  is  known.  The 
action  lines,  ad,  de,  and  ea,  of  three  forces  in  equilibrium 
and  the  magnitude  and  sense  of  one,  AD,  are  now  known.  To 
determine  the  remaining  elements,  we  have  only  to  draw  the 
force  triangle  for  the  three  forces,  ADEA.  Then  Z)£^  represents 
the  magnitude  and  sense  of  the  pull  of  the  chord,  and  EA  the 
magnitude  and  sense  of  the  pin  pressure. 

EXAMPLE. 

Suppose  that  the  bar  represented  in  fig.  98  is  supported  by- 
la  smooth  surface  at  A  (then  the  reaction  there 
must   be  horizontal)  and  by  a  pin  joint  at  B. 
Determine  the  supporting  forces  graphically. 

141.  Problem  III.  —  The  forces  of  a  non^ 
parallel  system  in  equilibrium  are  completely  known 
except  three  whose  lines  of  action  only  are  known. 
It  is  required  to  determine  completely  these  three.* 

Algebraic  Solution. — There  are  three  conditions  ^*^'  ^ 

of  equilibrium  for  a  system  of  this  kind,  as  in  the  preceding 

*  If  the  three  unknown  forces  are  concurrent  or  parallel,  the  problem 
is  indeterminate. 


200  lbs. 


100  lbs. 


126    APPLICATIOm  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI  ''\ 

article.     Either  set  of  equations  determines  the  unknowns,  the  \ 
magnitude  (and  sense)  of  three  fofees. 

Illustration:  Let  the  data  be  the  same  as  that  of  the  illus-  \ 
tration  of  the  preceding  article,  but  imagine  P  replaced  by  its  ! 
components  Px  and  Py  as  in  fig.  99.  j 

j^_^ 2^ 1— I^-r 2^—- r-1^-4---  ■ 


60  lbs.  100  lbs.      25  lbs. 
Fig.  99. 


(a)  Employing  the  first  set  of  equilibrium  equations,  . 

JF^^-Px  +  Qcosao^^o; (i;^ 

2'Fj/  =  Py  +  Q  sin  60°  — 50  — 100  — 25  =  0;     .     .     .     {2)\ 
2"  Ma  =  Q- 6  sin  60°  — 50-2 —  100-3  — 25-5=0.   •     •     {Z)\ 

Equation  (3)  is  just  like  eq.  (3)  of  the  preceding  article.  The  \ 
value  of  Q  determined  from  (3),  substituted  in  (i)  and  (2)  j 
leaves  them  with  two  unknowns,  Px  and  Py.  Now  it  is  con- J 
siderably  easier  to  solve  (i)  and  (2)  for  Px  and  Py  than  to  solve  j 
(i)  and  (2)  of  the  last  article  for  P  and  d,  as  may  be  seen  by  \ 
trial.  The  reason  lies  in  the  fact  that  both  sin  Q  and  cos  0  \ 
appear  in  those  equations,  and  one  really  has  to  solve  three  | 
equations,  the  third  one  being  sin^^  +  cos^^  =  i.  \ 

(b)  Employing  the  second  set  of  equilibrium  equations,  j 

IFx=-Px  +  Q  cos  6o°  =  o]       .......     (4)  j 

2^^  =  0-6  sin  6o°-5o-2-ioo. 3-25-5  =  0;  .     .     (5)] 
i'M5=-Py-6  +  5o-4  +  ioo-3  +  25-i=o.     .     .     .     (6)  i 

Equations  (5)  and  (6)  contain  but  one  unknown  each;  no  elimi-  \ 
nation  therefore  is  necessary  to  obtain  Py  and  Q.  The  value  of  ' 
Q  substituted  in  (4)  leaves  but  one  unknown  in  that  equation. 

(c)  The  third  set  of  equilibrium  equations  may  be  written  ) 
for  the  forces  of  this  example  so  that  but  one  unknown  will  j 
appear  in  each  equation,  and  then  no  elimination  is  necessary  ] 
in  solving  the  equations.     The  student  should  so  write  them.      j 

Graphical  Solutions — (i)  If  one  imagines  any  two  of  the; 
unknown  forces  replaced  by  their  resultant,  the  problem  is  \ 
transformed  to  Prob.  II  (art.  140).     Thus,  let  F\  F" ,  and  F'"  { 


§V.] 


THREE   TYPICAL  PROBLEMS. 


127 


denote  the  three  unknowns,  and  R  the  resultant  of  any  two  of 
them,  F"  and  F'"  say;  then  in  the  transformed  system  there 
are  two  partially  unknown  forces  F'  and  R.  The  action  line 
of  F'  is  known,  also  one  point  in  that  of  R,  the  intersection 
of  the  action  lines  of  F"  and  F'" .  Therefore,  by  the  method 
of  Prob.  II,  F'  and  R  may  be  determined,  and  then  R  may  be 
resolved  into  two  components  whose  action  lines  are  those  of 
F"  and  F'"  respectively;  these  components  are  F"  and  F'" . 

Illustration:  It  is  required  to  determine  the  supporting 
forces,  "reactions,"  on  the  overhanging  framework  of  fig.  100 
when  loaded  as  shown.  The  sup- 
ports are  such  that  the  reaction 
above  is  directed  along  the  rafter, 
and  reactions  below  are  horizontal 
and  vertical.  Imagine  the  lower 
two  reactions  compounded  and 
call  the  resultant  R',  its  line  of 
action  is  unknown,  but  it  passes 
through  the  lower  point  of  support. 
Then  the  system  in  equilibrium 
consists  of  R,  the  upper  reaction, 
and  the  loads.  The  force  polygon 
for  the  known  forces  is  ABODE, 
and  since  the  direction  of  the 
upper  reaction  is  known,  the  poly- 
gon can  be  continued  by  a  line 
from  E  parallel  to  the  upper  reaction, 
of  that  line  is  to  be  marked  F,  and  then  the  force  R  will  be 
FA,  since  the  force  polygon  is  to  be  closed. 

The  strings  which  intersect  on  af  are  oa  and  of  (art.  38),  and 
since  one  point  on  af,  P,  is  known  arid  also  OA,  the  string  oa 
may  be  drawn.  Continuing  the  construction,  ob  is  drawn  next, 
then  oc,  od,  and  oe.  Now  /  is  a  point  in  of  and  P  is  another; 
hence* 0/  may  be  drawn  and  then  the  corresponding  ray.  That 
ray  determines  F,  and  then  EF  and  FA  represent  the  upper 
reaction  and  R  respectively,  and  the  horizontal  and  vertical 
components  of  FA  (EG  and  GA)  represent  the  supporting 
.  forces  at  P. 


Fig.  100. 
The  as  yet  unknown  end 


128    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 


Scale:  l.in.= 3400  lbs. 


(2)  Find  the  resultant  of  the  wholly  known  forces;  then 
the  system  in  equilibrium  may  b*e  regarded  as  consisting  of 
that  resultant  and  the  three  partially  unknown  forces.  The 
special  condition  of  equilibrium  for  four  such  forces  is  that 
the  resultant  of  either  pair  balances  that  of  the  other  pair 
(art.  121). 

Illustration :  Let  the  data  be  the  same  as  that  of  the  pre- 
ceding illustration,  and  let  F  denote  the  upper  reaction  and 
H  and^  V  the  lower  ones  (fig.  loi).     The  resultant  of  the  loads 

is  a  force  of  1500  lbs.;  its  action  line 
is  ah.  Let  R'  denote  the  resultant  of 
the  pair  F,  1500,  and  R"  that  of 
the  pair  H,  V.  Now  R^  passes 
through  Q  and  R''  through  P,  and 
since  they  are  collinear,  PQ  is  the 
action  line  of  each.  Since  the  three 
forces  F,  1500,  and  i^"  are  in  equilib- 
rium, and  since  their  action  lines  and 
the  magnitude  and  sense  of  one  are 
known,  their  force  polygon  can  be 
drawn;  it  is  ABC  A,  BC  representing  the  magnitude  and  direc- 
tion of  F.  Since  the  four  forces,  1500,  F,  H,  and  V,  are  in 
equilibrium,  their  polygon  must  close;  hence  draw  lines  from 
A  and  C  parpllel  to  H  and  V  determining  D.  Then  ABCDA 
is  the  force  polygon  sought,  and  CD  and  DA  represent  the 
magnitudes  and  directions  of  V  and  H  respectively. 

EXAMPLES. 

I.  Suppose  the  truss  represented  in  fig.  102  to  be  supported 


8cale:Lln.=  16ft. 


Fig. 


on  a  smooth  surface  at  A  and  to  be  pinned  to  the  wall  at  B, 
Determine  the  reactions  at  A  and  B  due  to  the  loads,  by  both 
methods. 


§VL] 


JOINTED  FRAMES. 


129 


2.  Determine  the  reactions  on  the  crane  in  fig.  103  due  to 
the  load  and  weights  of  members  by  both  methods.     (The 


W^.  of  post      0.8  toa 
"    ""boom      .9   " 
"     "  "brace    1.1 


Fig.   103. 

supporting  surfaces  are  such  that  the  upper  reaction  is  hori- 
zontal and  below  one  is  horizontal  and  one  vertical.) 


§  VI.  Jointed  Frames. 

142.  Definitions. — A  plane  framework  is  one  all  members  of 
which  are  parallel  to  a  plane ;  only  such  frames  are  considered 
in  this  section.  A  jointed  frame  is  one  the  members  of  which 
are  fastened  by  pin-joints,  the  pin  being  perpendicular  to  the 
plane  of  the  frame.  It  is  assumed  in  the  following  that  the 
pins  are  smooth. 

143.  The  Pin  Pressures. — The  simplest  kind  of  a  member 
IS  one  which  is  straight  and  is  joined  to  others  at  its  ends  (fig. 
104).  .  Such  a  member,  if  sustaining  no  load,  is  subjected  to 


Fig.  104. 

three  forces,  its  weight  and  the  pin  pressures,  W,   P',  and  P" 
respectively.     Unless  the  member  is  vertical,  the  action  lines  of 


13°    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  Vl.j 

•I 
the  pin  pressures  do  not  coincide  with  the  axis  of  the  member;  for,  i 
assuming  that  they  do,  it  is  plairw.that  P'  and  P"  cannot  bal-i 
ance  W.  If  the  weight  of  the  member  is  neglected,  the  twoj 
forces  P'  and  P"  act  along  the  axis  and  are  equal;  but,  in\ 
general,  a  pin  pressure  is  not  directed  along  the  axis  of  the] 
member  on  which  it  acts.  / 

144.  General  Direction  for  Solving  Examples. — The  student  j 
should  read  again  art.  127.  In  the  following  examples  it  will^ 
not  always  be  evident  to  the  beginner  which  body  and  which , 
force  system  to  consider  for  determining  any  particular  force  J 
Often  several  may  be  selected  in  the  manner  mentioned  in  the; 
article  referred  to,  and  it  may  be  that  there  is  but  little  choice] 
between  them;  but,  as  a  general  rule,  the  body  should,  if  pos-| 
sible,  be  so  selected  that  the  number  of  unknown  elements  in^ 
the  external  system  of  forces  acting  upon  it  shall  not  exceed] 
the  number  of  conditions  of  equilibrium  for  that  system.  i 

EXAMPLES.  ] 

I.  Determine  the  forces  upon  each  member  of  the  crane  in] 
fig.  103,  neglecting  their  weights  and  taking  x  equal  to  16  ft.       \ 

Solutions:  (i)  Algebraic.  Fig.  105  represents  the  whole? 
crane,  its  members  and  groups  of  them,  and  the  corresponding! 
external  forces.  There  are  four  external  forces  applied  to  the! 
crane,  namely,  the  load,  and  the  reactions  X,  F,  and  H  (fig.i 
105a).  The  brace  (fig.  105^)  is  subjected  to  two  forces  only;j 
hence  they  are  collinear,  and  each  acts  along  AC.  The  boorrt- 
(fig.  105c)  is  subjected  to  three  forces, — one  at  C,  one  at  B;: 
and  the  load;  the  one  at  C  is  the  reaction  corresponding  to  the^ 
pressure  on  the  upper  end  of  the  brace,  and  hence  is  collinear i 
with  that  pressure.  The.  force  at  B  is  unknown  in  direction, J 
and  is  therefore  represented  by  its  components  Bx  and  ByS^ 
The  post  (fig.  io5<i)  is  subjected  to  five  forces, — X,  F,  H,  one| 
at  A ,  and  one  at  B ;  the  latter  two  are  reactions  corresponding! 
to  the  forces  upon  the  left  ends  of  the  boom  and  brace  respect-] 
ively.  \ 

An  examination  of  these  systems  reveals  several  orders  ofi 
procedure.  In  general,  it  is  advisable  to  determine  the  reac-| 
tions  on  the  entire  frame  first,  if  possible;  here  it  is  possible,! 


§vi.3 


JOINTED  FRAMES. 


131 


for  the  system  applied  to  the  entire  crane  is  coplanar  non- 
concurrent  non-parallel  with  three  unknowns.  Their  deter- 
mination is  left  for  the  student;  the  values  for  X,  Y,  and  H 
are  7.1 1,  8,  and  7.1 1   tons  respectively.     System  {c)  might  be 

-H 


TX 


4 


(e) 


X 

■■■-*.. 

1., 

- 

^«             4  Igl 

-<; 

(0) 

^^(d) 

cj/  .K 

/  ^ 

/ 

8* 

Fig. 


105. 


solved  next,  for  it  is  coplanar  non-concurrent  non-parallel 
with  three  unknowns.  The  student  should  prove  that  B^,  By, 
and  C  equal  respectively  10.67,  1.14,  and  14.05  tons.  From 
system  (6)  it  is  plain  that  A  equals  C.  Finally,  5  =  10.73, 
and  the  inclination  of  B  with  the  horizontal  is  tan~^^y/5x  = 
6°  7'.     All  the  unknowns  have  now  been  determined. 

The  student  should  examine  the  remaining  force  systems 
represented  in  fig.  105,  and  determine  other  orders  of  procedure 
for  determining  the  unknowns. 

(2)  Graphical.  The  system  of  external  forces  applied  to 
the  crane  may  be  solved  like  Prob.  Ill  (art.  141).  The  force 
polygon  for  the  four  forces  is  ABCDA  (fig.  io6a),  AB  repre- 
senting the  load.  The  forces  applied  to  the  boom,  being  three 
in  number,  are  concurrent;  hence  their  action  lines  are  known. 
Since  the  magnitude  of  one  force  is  known,  their  force  polygon 
can  be  drawn,  thus  determining  the  magnitudes  of  the  other 
two.  The  polygon  is  ABEA  (fig.  1066),  AB  representing  the 
load.     All  the  unknowns  have  now  been  determined. 


132     APPLICATIOm  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.   [Ch.  VI. 

Instead  of  solving  the  force  system  applied  to  the  boom,  we 
might  have  solved  that  applied  tg  the  mast.  There  are  five 
forces  in  that  system, — three  wholly  known  {BC,  CD,  and  DA), 
the  action  line  of  one,  and  a  point  in  that  of  the  last.  The 
resultant  of  the  three  wholly  known  forces  is  BA  (fig.  io6c); 


b 

c          / 

^/ 

a 

b 

//             D 

r    / 
/ 

/      a 

/ 
/ 

A 

d     d 

c 

a            f 

B 

(a) 


Scales,  of  space  diagram    1  ln.=  20  ft, 

"vector       "  lin.=  20tons- 

FiG.    io6. 


and  if  the  three  known  forces  be  imagined  replaced  by  their 
resultant,  then  the  system  consists  of  three  forces,  and  it 
must  be  concurrent.  The  action  line  of  the  last  force  is  now 
known  since  it  passes  through  the  action  lines  of  the  result- 


Load 


Fig.  107. 


Fig.  108. 


ant  and  the  fourth  force,  and  the  force  polygon  for  the  three 
forces    may  be    drawn,  thus    determining   the   remaining   un- 


§vr.j 


JOINTED  FRAMES. 


133 


knowns.     The  polygon  is  BAEB,  AE  and  KB  representing  the 
fourth  and  fifth  forces.* 

2.  The  weights  of  the  post  and  jib  of  the  crane  represented 
in  fig.  107  are  i  and  }  ton  respectively;  that  of  the  tie  may  be 
neglected.  Take  the  load  as  8  tons  and  ;i;  as  10  feet,  and  com- 
pute all  the  forces  on  each  member  by  both  methods. 

3.  The  weights  of  the  post  and  jib  represented  in  fig.  108 
are  1200  and  1500  lbs.  respectively;  that  of  the  tie  may  be 
neglected.  For  a  load  of  4  tons  compute  all  the  forces  upon 
each  piece  by  both  methods.  Disregard 
counterweight,  shown  dotted. 

4.  Solve  the  preceding  example  on 
the  supposition  that  the  crane  has  a 
counterbalance  whose  weight  is  10  tons, 
its  centre  of  gravity  being  9  ft.  from 
the  axis  of  the  post. 

5.  Determine  all  the  forces  upon 
each  member  of  the  crane  repre- 
sented in  fig.  109  due  to  a  load  of  3 
tons  7  ft.  from  C.  It  is  impossible  to 
analyze  this  crane  by  the  preceding  principles.  To  make  it 
possible,  suppose  that  the  hole  in  the  mast  at  C  is  slotted  as 
shown  (then  the  pressure  there  is  vertical),  and  that  there  is  no 
member  AE. 

6.  Fig.  iioa  represents  a  type  of  hydraulic  crane.  The 
plunger  works  inside  a  hollow  mast  and,  pressing  against  the 


Fig. 


109. 


*  The  solution  of  ex.  i  suggests  these  general  directions  for  "analyz- 
ing" cranes: 

(i)  Make  a  sketch  of  the  entire  crane,  and  represent  as  far  as  possible 
all  the  external  forces  acting  upon  it. 

(2)  Apply  the  proper  conditions  of  equilibrium  •  (art.  125)  to  such 
external  forces  and  determine  as  many  of  the  unknown  elements  as 
possible. 

(3)  Make  sketches  of  members  or  collections  of  members  of  the 
crane,  and  represent  as  far  as  possible  all  the  external  forces  acting 
upon  them.  (In  this  connection,  bear  in  mind  art.  143  and  the  "law 
of  action  and  reaction.") 

(4)  Inspect  the  force  systems  of  the  sketches,  noting  especially 
the  unknown  forces  in  each.  Careful  inspection  will  suggest  how  to  deter- 
mine the  unknowns. 


134     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.   [Ch.  VI.  ] 

] 

bottom  of  the  boom,  raises  and  lowers  the  load  and  boom  to-  'I 

gether.     Compute  the  forces  upon  each  piece  due  to  a  load  of  j 

lo  tons,  X  feet  from  the  axis  of  the  plunger  when  the  pin  A  is  ^ 
y  feet  above  the  floor. 

Solution:  Consider  first  the  entire  crane  (fig.  nob),  except  i 

post  and  plunger.     The  external  forces  applied  to  it  consist  of  \ 

the  load,  the  post  pressure  upon  the  upper  roller,  R^,  the  post  \ 

pressures  on  the  lower  rollers,  R^  each,  and  the  plunger  pres-  \ 

sure  P.     The  three  unknowns  may  be  determined,  for  the  sys-  \ 


Fig.   iio. 

tem   is   coplanar   non-concurrent    non-parallel.      The    solution 
gives 

Ri  =  i,4$x,    R2  —  0.JIX,    and    P=  10  tons. 

To  continue,  we  may  consider  next  the  pin  and  roller  at  B, 
The  external  forces  upon  this  body  are  five  in  number,  the  pres- 
sure on  the  roller  i  .43^:,  the  forces  exerted  on  the  pin  by  the  two 
members  AB  and  the  two  BC,  The  four  pin  pressures  are 
directed  along  the  axes  of  corresponding  members  (fig.  hoc). 
Solution  of  the  equations  of  equilibrium  of  this  system  gives 

F'  =  o.26^    and     F^'^o.S^x  tons. 

Upon  the  pin  at  A  there  are  applied  five  forces,  two  by  the 


^  VI.] 


JOINTED  FRAMES. 


135 


members  AB,  two  by  the  rollers,  and  one  by  the  boom;  their 
magnitudes  are  respectively  0.26^,  0.71:*:,  and  i.()']X.  The 
student  should  prove  the  values  given. 

7.  Include  the  weights  of  the  members  in  ex.  i  and  solve. 
(Notice  that  the  forces  applied  to  the  ends  of  the  brace  are  not 
directed  along  its  axis.  It  will  be  convenient  in  an  algebraic 
solution  to  replace  each  unknown  force  whose  direction  is  un- 
known by  its  horizontal  and  vertical  components.) 

8.  The  frame  of  fig.  1 1  i(a) 
rests  upon  smooth  surfaces  at 
A  and  B.  Determine  the 
forces  (pin  pressures)  upon 
each  member. 

Solution :  From  a  conside- 
ration of  the  external  forces  on 
the  collection  of  bars  it  follows 
readily  that  the  forces  at  A 
and  B  are  446  and  554  lbs.  re- 
spectively. Fig.  Ill  {h),  (c), 
and  (d)  represents  the  external 
system  on  each  member,  each 
pin  pressure  being  replaced  by 
its  horizontal  and  vertical 
components.  No  unknown 
of  the  system  (b)  or  (c)  can 
be      determined      from     the 

equilibrium  equations  of  the  system,  but  the  forces 
and  Ey  may  be  computed  from  the  equations  for  system  (d). 
Making  use  of  these  values,  in  (a)  and  (b),  the  remaining 
unknowns  can  be  determined.  The  student  should  make  the 
determination. 

9.  The  bars  of  the  frame  of  fig.  112(a)  are 'uniform,  AC,  BC, 
and  AB  weighing  150,  100,  and  200  lbs.  respectively.  Com- 
rute  the  forces  upon  each  bar. 

Solution:  The  forces  applied  to  each  member  are  repre- 
sented in  fig.  112  {b),  {c),  and  {d),  each  force  whose  direction  is 
unknown  being  represented  by  its  horizontal  and  vertical  com- 
ponents. The  senses  of  the  vertical  components  at  C  are  not 
obvious,  so  they  are  assumed  (see  foot-note,  p.  120). 

From  a  consideration  of  the  external  system  on  the  entire 


D. 


X36    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

frame,  R^  and  R^  are  readily  found  to  be  229  and  221  lbs.  re- 
spectively. From  the  system  {d),  4  2/  and  B y  are  iornid  to  he 
129  and  121  lbs.  respectively.  Supplying  the  value  of  Ay  in 
system  (6)  or  By  in  system  (c),  the  remaining  unknowns  may 
be  computed. 


Instead  of  following  the  order  above,  one  might  write  the 
equilibrium  equations  for  systems  (6)  and  (c)  and  solve  them 
for  the  six  unknowns,  and  lastly  determine  i^j  and  R2. 


§  VII.     Jointed  Frames  (Continued). 

145.  Kind  of  Frames  Considered. — The  jointed  frames  con- 
sidered in  this  section  differ  from  those  of  the  preceding  section 
in  construction  and  in  loading.     It  is  assumed  that 

(a)  each  member  connects  ofily  two  joints, 

(b)  each  load  is  applied  so  that  its  action  line  passes  through  the 
axis  of  a  joint. 

146.  "Force  or  Stress  in  a  Member." — Fig.  113(a)  repre- 
sents a  member  of  such  a  frame  as  described  in  the  preceding 
article,  under  two  loads  U  and  U\  The  pin  pressures  are 
denoted  by  P'  and  P",  and  the  weight  of  the  member  is 
neglected  or  not  considered.     Let  the  resultants  of  the  forces 


§VIL]  JOINTED  FRAMES.  I37 

at  the  left  and  right  ends  be  denoted  by  R'  and  R"  respect- 
ively; the  action  line  of  each  passes  through  the  centre  of 
the  corresponding  hole.  Since/?'  and  i?"  balance,  they  must 
be  collinear,  and  their  action  lines  must  coincide  with  the  axis  of 
the  member. 

Now  any  two  parts  of  the  member,  as  M  and  A/",  exert  forces 
on  each  other,  and  the  lines  of  action  of  those  forces  coincide 

\-       \ 


<f 


4-^     M      h»  -<-r  N         -^  (b) 


^'^'         M        -K-    ->|  N  4-^"  f  o) 


Fig.  113. 

with  the  axis  of  the  member.  For,  the  force  which  M  exerts  on 
N  balances  R" ,  therefore  its  action  line  must  coincide  with  that 
of  R"',  also  the  force  which  N  exerts  on  M  balances  R\  there- 
fore its  action  line  must  coincide  with  that  of  R' . 

Observe  carefully  the  relations  in  fig.  113  (6)  and  {c). 
In  (6),  the  forces  at  the  section  between  M  and  N  are  pulls, 

or  the  stress  is  tensile  (art.  104),  and  the  member  is  stretched 

by  R'  and  R" . 
In  (c),  the   forces    at    the    section    between    M  and    N    are 

pushes,  or  the  stress  is  pressural  (art.  104),  and  the  member 

is  compressed  by  R'  and  R" . 

By  force  or  stress  in  a  member  is  meant  either  of  the  forces 
which  a  part  of  it  exerts  upon  the  other.  The  examples  in  this 
section  relate  to  the  determination  of  the  forces  in  the  members 
of  jointed  frames. 

147.  Method  for  Determining  the  Force  or  Stress  in  a  Member. 

(i)  Determine  the  reactions  on  the  frame,  or  truss. 

(2)  Imagine  the  truss  separated  into  two  distinct  parts  *  so 

*  Such  division  of  a  truss  is  also  described  as  ""passing  a  section,"  the 
term  section  referring  to  the  imaginary  surface  of  separation. 


138    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

that  the  member  under  consideration  is  one  of  the  members 
separated,  and  so  that  the  unknown»elements  in  the  system  of 
external  forces  applied  to  one  part  of  the  truss  does  not  exceed 
the  number  of  algebraic  conditions  of  equilibrium  for  that 
system. 

(3)  Apply  the  appropriate  condition  or  conditions  of  equi- 
librium necessary  to  determine  the  desired  force. 

The  student  should  bear  in  mind  that  the  system  of  external 
forces  with  reference  to  any  part  of  a  truss  consists  of  the  loads 
and  reactions  applied  to  that  part  and  the  forces  which  the 


Fig.   114. 

other  part  exerts  upon  it.  These  latter  forces  are  exerted  upon 
the  ' '  cut ' '  ends  of  the  members  belonging  to  the  part  considered 
and  are  exerted  along  their  axes. 

Illustration  i. — It  is  required  to  determine  the  force  in 
the  member  AB  of  the  truss  of  fig.  1 14(a). 

The  reactions  are  supposed  to  have  been  determined.  If 
the  section  cutting  AB  be  passed  as  at  i .  the  external  system 
©neither  the  upper  or  the  lower  part  of  truss  (fig.  1146)  includes 
four  unknown  forces,  those  in  the  four  members  cut.  Now 
the  former  system  is  concurrent  and  since  it  has  only  two  con- 
ditions  of   equilibrium   the   unknowns   cannot   be   determined 


§  VIL]  JOINTED  FRAMES.  139 

from  that  system.  The  system  on  the  lower  part  is  non-con- 
current, and  as  it  has  only  three  conditions  the  unknowns  can- 
not be  determined  from  it. 

If  the  section  be  passed  as  at  II,  the  system  on  each  part  of 
the  truss  contains  only  three  unknown  forces  (fig.  114c);  and 
since  each  system  is  non- concurrent,  there  are  three  conditions 
of  equilibrium,  and  the  force  desired  can  be  obtained  from  either 
system. 

Illustration  2. — It  is  required  to  determine  the  stress  in 
member  BC ,  fig.  114(a). 

No  matter  where  the  section  cutting  CB  is  made,  the  ex- 
ternal system  on  either  part  will  contain  more  unknown  forces 
than  the  number  of  conditions  of  equilibrium  for  the  system, 
and  the  desired  stress  cannot  be  so  directly  determined.  Thus, 
if  the  section  is  made  as  at  ///,  the  system  on  the  left  part  is  as 
represented  in  fig.  114(6^),  and  there  are  four  unknowns,  the 
forces  in  the  four  cut  members.  If  now  the  force  in  CD,  BE, 
or  EF  can  be  determined,  its  value  may  be  supplied  in  fig.  i  i^{d) 
and  the  system  can  then  be  solved  for  the  desired  force,  for  there 
are  three  conditions  of  equilibrium  and  but  three  unknowns. 
The  force  in  CD  can  be  determined  from  the  system  of  fig. 
114(0. 

When  the  stress  in  each  member  of  a  truss  is  required,  a 
certain  order  of  determining  them,  depending  on  the  case  in 
hand,  is  more  convenient  than  others;  but  in  all  cases,  the  sec- 
tions are  made  according  to  direction  (2)  as  stated  above.  It 
does  not  fall  within  the  scope  of  this  book  to  explain  fully  the 
most  convenient  orders  of  procedure  for  the  different  cases. 
A  good  general  method  is  to  make  the  sections  so  that  the  ex- 
ternal system  on  one  of  the  parts  of  the  truss  shall  be  simple, 
containing  few  unknowns  and  easy  to  solve.  This  matter  is 
partially  illustrated  in  the  solution  of  the  first  of  the  following 
examples  and  in  arts.  148-151. 

EXAMPLES. 

I.  Determine  the  force  in  each  member  of  the  frame  of 
fig-  ii5(^)  ^iie  to  the  load  of  1000  lbs 


I40    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

Solution :  Imagine  the  truss  divided  into  two  distinct  parts 
as  shown  in  fig.  115  (b)  and  (c).  »The  external  forces  upon 
the  left  part  are  400  lbs.,  F/  and  F/;  upon  the  right  part  600 
lbs.,  1000  lbs.,  F/'  andF/' *  Each  is  a  system  in  equilibrium, 
and  the  unknowns  may  be  determined  from  the  equilibrium 
equations  for  either  system. 

lOOOjlbs. 


1000  lbs. 


^> 


^^ 


(gr) 


f  400  lbs. 


■Fig.   115. 


To  determine  the  force  in  the  member  BC,  the  truss  should 
be  imagined  as  separated  into. two  parts  so  that  BC  is  one  of 
the  members  cut,  for  example  as  in  fig.  115  (d)  and  (e).  The  ex- 
ternal forces  on  the  left  part  are  400  lbs.,  1000  lbs.,  F/  and  F3'; 
upon  the  right  part  600  lbs.,  Fg"  ^^^  -^3"-  Each  is  a  system  in 
equilibrium,  and  the  equilibrium  equations  for  either  determine 
the  remaining  unknown  F3. 


*F'  —  F"    F  '  —  F 


etc. 


§VII.]  JOINTED  FRAMES.  14* 

In  fig.  115  (/)  and  fe),  there  are  represented  the  external 
systems  on  the  parts  of  the  truss  made  by  cutting  members  AC 
and  BC.  The  equilibrium  equations  for  either  determine  F^ 
and  F3. 

Of  course,  consideration  of  all  the  systems  represented  is 
not  necessary  for  a  solution.  They  are  referred  to  here  merely 
to  show  that  the  solution  may  be  made  in  several  different 
ways.  One  of  these  ways  is  by  means  of  the  systems  of  fig. 
115  {e)  and  (/),  which  may  be  carried  out  thus: 

Equilibrium  equations  for  system  {e)  are 

2'F,  =  F/'cos36°52'-i^2''  =  o; 
IFy=  -F/'  sin  36°  52'  +  6oo  =  o;  * 

hence 

F3"  =  1 000     and     F^'  =  800  lbs. 

Having  determined  F/\  F3'  is  known;  it  is  a  push,  that  is,  it 
acts  upward  (fig.  115/)  and  its  value  is  1000  lbs.  Only  one 
unknown  remains  in  system  (/),  and  the  following  equilibrium 
equation  suffices  for  its  determination: 

IFx  =  F^"  cos  26°  34'  —  1000  cos  36°  52^  =  0; 
hence  F/'  =  894. 5  lbs. 

Supposing  the  senses  of  the  unknown  forces  in  the  two 
systems  just  considered  not  apparent,  and  following  the  sug- 
gestion of  the  foot-note,  the  first  two  equations  above  become 

-F3-cos36°52'-F/'  =  o; 
Fg"  sin  36°  52' +  600  =  0; 

from  which         F3"  =  — 1000     and     F^"  =  +  800  lbs. 

*  In  simple  trusses  the  kind  of  stress  in  any  member  is  apparent. 
F©r  example,  in  Fig.  115(a),  AC  and  BC  are  in  compression  and  AB  in 
tension;  then  F^  and  F3  are  pushes  and  F2  is  a  pull.  When  the  senses  of 
the  forces  are  not  apparent,  we  may  follow  the  suggestion  in  the  foot-note, 
p.  120,  but  it  is  convenient  to  always  assume  the  force  to  he  a  pull.  Then, 
according  to  the  foot-note,  the  force  is  actually  a  pull  or  push  {and  the  member 
is  in  tension  or  co7npression)  according  as  its  computed  value  is  positive  or 
-negative. 


142     APPLICATION'S  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. ^ 


Interpreting  these  signs  in  accordance  with  the  foot-note,  F^" 
is  a  push  and  F/'  a  pull,  i.e.,  BC*is  in  compression  and  AB  in 
tension — results  agreeing  with  the  first  solution. 

2.  Determine   the   force   in   each   member  of  the  truss   of 


fig.   1 1 6. 


Ans.  AF,  i6oo  lbs.  tension. 


Fig.  ii6. 


Fig.  117. 


3.  Determine     the    force    in    each 
member  of  the  truss   of  fig.  117. 

Ans.  AD,  2600  lbs.  compression; 
AC,  1300  lbs.  tension. 

4.  Determine     the    force    in    each 
B  member   of  the   truss   of  fig.   118. 

5.  Determine     the    force    in    each 
member  of  the  truss  of  fig.  102,  it  being 

supported  as  there  described. 

148.  Graphical  Method  for  "Analyzing  Trusses." — Graph- 
ical methods  are  especially  well  adapted  for  solving  problems 
like  the  preceding.  As  in  the  algebraic  method,  the  truss  is 
imagined  separated  into  two  parts  and  then  the  attention  is 
directed  to  the  external  forces  acting  upon  either  part.  Graphi- 
cal instead  of  algebraical  conditions  of  equilibrium  are  then 
applied  to  the  system  of  forces  to  determine  the  unknowns. 
In  making  the  imaginary  separations  of  the  truss,  care  should 
be  taken  to  cut  not  more  than  three  members,  the  forces 
in  which  are  unknown.*  It  is  advantageous  to  make  the  separa- 
tion so  that  not  more  than  two  such  members  are  cut.  If  that 
be  done,  a  single  force  polygon  will  determine  the  two  unknowns, 
while  if  three  be  cut,  a  force  polygon  and  a  funicular  polygon, 
or  the  equivalent,  are  necessary  to  determine  the  unknowns. 

*  These  members  must  not  meet  at  the  same  joint. 


§VII.]  JOINTED  FRAMES.  I43 

149.  Notation.— The.  notation  described  in  art.  11  when 
applied  in  the  graphical  analysis  of  trusses  can  be  advanta- 
geously systemized  as  follows.  Each  triangular  space  in  the 
truss  diagram  is  marked  by  a  small  letter,  also  the  space 
between  consecutive  action  lines  of  the  loads  and  reactions 
(see  fig.  119)  Then  the  two  letters  on  opposite  sides  of  any 
line  serve  to  designate  that  line,  and  the  same  large  letters  are 
used  to  designate  the  magnitude  of  the  corresponding  force. 
This  scheme  of  notation  is  a  great  help  in  graphical  analyses  of 
trusses. 

Illustration. — Determine  the  force  in  each  member  of  the 
truss  of  fig.  119. 

Solution:  Evidently  the  reactions  each  equal  one-half  the 
load,  or  2000  lbs.     Imagine  the  truss  separated  into  two  parts,  as 

\i 

.IV 


II.  t>r  , 


-II  >C/vS 


ft 


Fig.   119. 


by  the  arc  /.  The  external  forces  upon  the  left  part  are  repre- 
sented as  far  as  known  in  fig.  120(a);  since  they  are  in  equilib- 
rium, their  polygon  closes,  and  in  constructing  it,  the  unknowns 
will  be  determined.  Beginning  with  the  knowns,  AB  is  drawn 
to  represent  2000  lbs.,  BC  to  represent  500  lbs.;  and  then  a 
line  from  A  (or  C)  parallel  to  the  action  line  of  one  of  the  un- 
knowns and  a  line  from  C  {or  A)  parallel  to  the  other  are  drawn. 
The  last  two  lines  determine  D  (or  D'),  and  the  closed  polygon 
is  ABCDA  (or  ABCD'A)\  hence  the  forces  in  the  members  cd 
and  ad  are  represented  by  CD  and  DA  (3000  and  2600  lbs.) 
respectively.  From  the  force  polygon,  it  is  seen  that  CD  is  a 
push,  and  DA  is  a  pull;  hence  the  members  cd  and  ad  are  in 
compression  and  tension  respectively. 

We  may  next  imagine  the  truss  separated  into  two  parts 
as  by  //  or  //'  (fig.  119);   in  either  case,  there  are  but  two  un- 


144     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VL 

known  forces  in  the  external  system  applied  to  either  part.  If 
we  choose  the  part  within  //,  we,^have  the  simpler  system  to 
deal  with;  the  forces  of  it  are  represented  in  fig.  120(6)  as  far 
as  known.  The  force  polygon  may  be  drawn  thus:  DC  to 
represent  3000  lbs.,  CF  to  represent  1000  lbs.,  a  line  from  F 
parallel  to  one  of  the  unknowns  and  one  from  D  parallel  to  the 
other.     The  last  two  lines  determine  E,  and  the  force  polygon 


is  DCFED ;  hence  the  forces  in  the  members  de  and  ef  are  repre- 
sented by  £Z)  and  FE  (866  and  2500  lbs.)  respectively.  Both 
members  are  in  compression. 

We  next  imagine  the  truss  separated  into  parts  as  by  ///. 
Choosing  the  part  within  ///,  we  have  a  simple  system  to  deal 
with ;  the  forces  of  it  are  represented  as  far  as  known  in  fig.  1 20(c) . 
Their  force  polygon  may  be  drawn  thus:  AD  to  represent  2600 
lbs.,  DE  to  represent  866  lbs.,  a  line  from  E  parallel  to  one  of 
the  unknowns,  and  a  line  from  A  parallel  to  the  other.  The 
last  two  lines  determine  G,  and  the  force  polygon  is  ADEGA ; 
hence  the  forces  in  the  members  eg  and  ag  are  represented  by 


f  Of    THE  A 

(   UNIVERSITY    ) 


^^^fiikK 


VIL] 


JOINTED  FRAMES. 


145 


EG  and  GA   (866  and   1732  lbs.)  respectively.     Each  member 
is  in  tension. 

On  account  of  the  symmetry  of  the  truss  and  loading,  the 
forces  in  the  remaining  raembers  are  now  known. 

150.  Polygon  for  a  Joint. — In  drawing  the  force  polygon 
for  all  the  external  forces  on  the  part  of  a  truss  included  within 
a  small  circle  struck  from  a  joint,  it  will  be  advantageous  to 
represent  the  forces  in  the  order  in  which  they  occur  about  the 
joint. 

A  force  polygon  so  drawn  will  be  called  a  polygon  for  the 
joint;  and  for  brevity,  if  the  order  taken  is  clockwise,  the  poly- 
gon will  be  called  a  clockwise  polygon,  and  if  counter-clockwise 
it  will  be  called  a  counter-clockwise  polygon.  ABC  DA  (fig. 
1 20a')  is  a  clockwise  polygon  for  joint  /  of  fig.  119;  ABCD'A 
is  a  force  polygon  for  the  "forces  at  joint  /,"  but  it  is  not  a 
polygon  for  the  joint,  because  the  forces  are  not  represented  in 
the  polygon  in  the  order  in  which  the  forces  occur  about  the  joint. 

The  student  should  draw  the  counter-clockwise  polygon 
■for  the  joint  and  compare  with  ABCDA. 

151.  Stress  Diagrams. — If  the  polygons  for  all  the  joints  of 
a  truss  are  drawn  separately  as  in  the  illustration  in  art.  149, 
then  the  stress  in  each  member  will  have 
been  represented  twice.  It  is  possible 
to  combine  the  polygons  so  that  it  will 
not  be  necessary  to  represent  the  stress 
in  any  member  more  than  once,  thus 
reducing  the  number  of  lines  to  be 
drawn.  Such  a  combination  of  force 
polygons  is  called  a  stress  diagram. 

Figure  121  is  a  stress  diagram  for 
the  truss  of  fig.  119  loaded  as  there 
shown.  Comparing  the  par.t  consisting 
of  solid  lines  with  figs.  120  (a'),  (6'), 
and  {c'),  it  is  seen  to  be  a  combination 
of  the  latter  three  figures.  It  will  also 
be  observed  that  the  polygons  are  all 

clockwise  polygons,  but   counter-clockwise   polygons   could  be 
combined  into  a  stress  diagram. 


Fig.  121. 


146    APPLlCATlOm  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.   [Ch.  VI 

To  construct  a  stress  diagram  for  a  truss  under  given  loads: 
(i)  Letter  the  truss  diagram  as  directed  in  art.  149. 

(2)  Determine  the  reactions. 

(3)  Construct  a  force  polygon  for  all  the  external  forces  applied 

to  the  truss  (loads  and  reactions),  representing  them  in 
the  order  in  which  their  application  points  occur  about 
the  truss,  clockwise  or  counter-clockwise.* 
(4;  On  the  sides  of  that  polygon,  construct  the  polygons  for 
all  the  joints.  They  must  be  clockwise  or  counter-clock- 
wise ones  according  as  the  polygon  for  the  loads  and 
reactions  was  drawn  clockwise  or  counter-clockwise. 
(The  first  polygon  drawn  must  be  for  a  joint  at  which  but 
two  members  are  fastened;  the  joints  at  the  supports  are 
usually  such.  Next  that  joint  is  considered  (and  its  poly- 
gon is  drawn)  at  which  not  more  than  two  stresses  are 
unknown,  that  is,  of  all  the  members  fastened  at  that 
joint  the  forces  in  not  more  than  two  are  unknown.  Then 
the  next  joint  at  which  not  more  than  two  stresses  are 
unknown  is  considered ;  etc.f) 

These  directions  are  illustrated  in  the  following  solutions. 

EXAMPLES. 

I.  Solve  ex.  4  of  art.  147  by  the  graphical  method. 

Solution:  Supposing  the  reactions  to  have  been  deter- 
mined, we  draw  the  force  polygon  for  the  loads  and  reactions 
ABCDEFA  (fig.  1226);  it  is  a  clockwise  polygon.  We  may 
begin  by  drawing  the  clockwise  polygon  for  joint  I  or  II  \  for  the 
former  it  is  FABGF.X     Member  hg  is  therefore  in  compression 


*  The  part  of  that  polygon  -representing  the  loads  is  called  a  load  line. 

t  In  some  trusses,  after  the  polygons  for  a  few  joints  are  drawn,  there 
remains  no  joint  at  which  there  are  but  two  unknown  stresses;  fig.  123 
represents  such  a  one.  The  solution  of  ex.  5  explains  several  ways  of 
procedure  in  such  cases. 

X  The  student  is  urged  to  make  sketches  of  the  bodies  (parts  of  truss) 
upon  which  the  forces,  whose  polygons  are  being  drawn,  act.  A  force 
acting  upon  the  "cut"  end  of  a  member  and  toward  the  joint  is  a  push, 
and  the  stress  in  the  member  is  compressive;  if  it  acts  away  from  the 
joint,  it  is  a  pull  and  the  stress  is  tensile. 


iJVlL] 


JOINTED  FRAMES. 


U7 


;and  gf  in  tension.     Next  we  may  draw  the  clockwise  polygon 
[for  joint  II ,  III ,  or  IV \  for  the  first  it  is  CDEHC.     Member  ch 


(b) 


lin.=  4000  lbs. 


Iin.=  l6ft. 

Fig.   122. 


1  in. =4000  lbs. 


is  in  compression  and  eh  in  tension.  For  joint  ///,  the  polygon 
is  HEFGH  and  member  gh  is  in  tension.  If  the  work  has  been 
correctly  and  accurately  done,  the  line  GH  is  parallel  to  gh. 

2.  Solve  ex.  2  of  art.  147  graphically. 

3.  Solve  ex.  3  of  art.  147  graphically. 

4.  Solve  ex.  5  of  art.  147  graphically. 

5.  Analyze  the  truss  of  fig.  123  under  the  loads  shown. 
Solution :  Evidently  each  reaction  equals  one-half  the  whole 

load.  ABCDEE'D'C'B'A'FA  is  a  clockwise  polygon  for  the 
loads  and  reactions.  The  polygon  for  joint  i  may  be  drawn 
first;  it  is  FABGF.  Next,  that  for  joint  2  may  be  drawn; 
it  is  GBCHG.  Then  that  for  joint  3  may  be  drawn;  it  is 
FGHIF.  The  polygons  for  joints  i',  2',  and  3'  are  FA'B'G'F, 
G'B'C'H'G\  and  FG'HTF  respectively. 

No  joint  remains  at  which  there  are  but  two  unknown  forces, 
and  no  more  polygons  can  be  drawn.  If  in  any  way  the  number 
of  unknown  forces  at  a  joint  can  be  reduced  to  two,  the  polygon 
for  that  joint  can  be  drawn  and  the  stress  diagram  can  be  com- 
pleted. There  are  several  ways  of  making  that  reduction. 
For  example,  if  the  force  in  ij,  jm,  or  mf  were  known,  the  poly- 
gon for  joint  4  could  be  drawn,  then  that  for  5,  6,  7,  and  8. 

The  force  in  mf  may  be  determined  by  "  passing  a  section" 
as  at  /  and  solving  the  external  system  on  either  part  of  the 
truss  for  the  desired  force.  The  system  consists  of  the  loads 
and  reaction  on  that  part  and  the  forces  in  the  members  cut. 


148    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

The  »  /stem  may  of  course  be  solved  graphically  or  algebrai- 
cally i  but  in  this  truss  the  algebraic  solution  is  much  the  sim- 
pler.    A.  moment  equation  for  either  system  with  joint  8  as 


Scale:  1  ln.=  5000  lbs. 
Fig.  123.  I 

origin  furnishes  the  value  of  the  force  in  mf  readily;  the  value  j 
is  3420  lbs.  and  the  stress  is  tensile.  \ 

This  force  in  fm  may  now  be  represented  in  the  proper  place  j 
in  the  stress  diagram  determining  M,  and  then  the  polygon  for  j 
joint  4  can  be  drawn;  it  is  MFIJM.  The  student  should  pick  j 
out  the  polygons  for  the  remaining  joints  and  determine  the  \ 
kind  of  stress  in  each  member.  \ 

There  are  other  ways  of  meeting  the  difficulty  presented  in  1 
this  form  of  truss,  but  that  here  given  is  the  most  general  ; 
and  can  be  applied  readily  to  other  forms. 

6.  Analyae  the  truss  represented  in  fig.  114(a)  under  the  ■ 
loads  shown. 


§  VIIL] 


ROUGH  SUPPORTS;   FRICTION. 


149 


VIIL     Rough  Supports;  Friction. 


Fig. 


124. 


152.  Definitions. — It  is  a  fact  of  experience  that  when  one 
body  slides  or  tends  to  slide  over  another,  the  slid- 
ing of  the  first  is  opposed  or  resisted  by  the  second. 
Thus,  suppose  that  fig.  124  represents  a  block 
which  slides  or  tends  to  slide  over  another  body 
towards  the  right;  the  second  body  exerts  some 
such  force  as  R  upon  the  block. 

The  force  which  one  body  exerts  upon  another  which  slides 
or  tends  to  slide  over  the  first  is  called  the  total  resistance;  it 
will  be  denoted  by  R.  The  component  of  the  total  resistance 
along  the  (plane)  surface  of  contact  is  called  sliding  resistance y 
or  more  commonly  friction;  the  component,  normal  to  the 
surface  is  called  normal  pressure.  (If  the  surface  of  contact 
of  the  two  bodies  is  not  plane,  the  force  exerted  at  each  ele- 
mentary part  of  the  surface  is  the  total  resistance  applied  to 
that  element,  and  its  components  in  and  normal  to  the  element 
are  the  friction  and  the  normal  pressure  applied  to  the  element.) 

Friction  is  called  kinetic  or  static  according  as  sliding  does 
or  does  not  take  place.  Static  friction  only  is  here  considered 
(kinetic  friction  is  discussed  later). 

Suppose  that  the  block  represented  in  fig.  125  weighs  10  lbs.. 


that  it  is  subjected  to  a  horizontal  pull,  P,  and  the  rubbing 
surfaces  are  such  that  P  must  exceed  6  lbs.  to  start  the  block. 
Fig.  125(a),  (6),  and  {c)  represent ^the  forces  acting  upon  the 
block  when  P,  as  it  increases,  reaches  values  of  2,4,  and  6  lbs. 
respectively.  Since  the  block  is  at  rest,  the  friction  at  the  three 
stages  equals  2,4,  and  6  lbs.  and  in  all  stages  N  equals  10  lbs. 
When  P  reaches  6+  lbs.  the  block  will  move  and  the  kinetic 


150    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

friction  would  be  something  less  than  6  lbs.,  as  has  been  dis- 
covered experimentally.  For  any  two  bodies  then,  the  fric- 
tion may  have  many  values  depending  on  whether  slipping 
occurs  or  not,  and  if  not,  on  how  great;  the  tendency  to 
slipping  is. 

The  friction  corresponding  to  impending  motion  is  called 
limiting  friction;  it  will  be  denoted  by  F' .  Evidently  limiting 
friction  is  the  maximum  value  of  the  friction  corresponding  to 
any  given  normal  pressure,  see  fig.  125  (a),  (6),  and(<:).  Limit- 
ing friction  has  been  studied  experimentally  and  many  impor- 
tant results  have  been  thus  r'>;duced. 

153.  The  Coefficient  of  Static  Friction  for  two  rubbing  sur- 
faces is  the  ratio  of  any  normal  pressure  between  the  surfaces 
and  the  corresponding  limiting  friction;  if  it  is  denoted  by  /, 

f  =  F'/N,     or     F'  =  fN. 

154.  The  Angle  of  Friction  for  two  rubbing  surfaces  is  the 
angle  between  the  directions  of  the  normal  pressure  and  the 
total  resistance  when  motion  is  impending.  Denoting  it  by 
cj)  (see  fig.  125c), 

tan  (j>  =  F'/N\     hence     tan  4>  =  f. 

155.  Angle  of  Repose. — If  a  block  be  placed  upon  an  in- 
clined plane,  the  inclination  at  which  slipping  would  be  impend- 

l\  ing  is  called  the  angle  of  repose  for  the  two  rub- 
bing  surfaces;    it  will  be  denoted   by   a.      From 
fig.   126    (representing  a   body  on   an  incline,  the 
angle  being  that  of  repose),  and  the  equations  of 
126.      equilibrium  for  the  forces, 

F'  =  W  sin  a     and     N  =  W  cos  a ; 

nence  tan  a  =  F'/N. 

Since  F'/N  =  f  =^ tan  (j),     a  =  (j),     and     tana=/; 

that  is,  the  angle  of  repose  for  two  surfaces  equals  their  angle  of 
friction,  and  the  tangent  of  the  angle  of  repose^equals  the  coeffi- 
cient of  friction. 


§VIIL]  ROUGH  SUPPORTS;    FRICTION.  151 

156.  Laws  of  Friction. — The  following  laws  relate  to  ''  solid 
friction ' '  (friction  between  solids)  and  are  based  entirely  on 
experiment. 

1.  The  coefficient  of  friction  for  two  surfaces  depends  upon 
the  nature  of  the  surfaces.  Thus  the  coefficient  varies  with 
the  materials,  with  the  smoothness  of  the  surfaces,  and  with 
the  lubricant,  if  any  is  used. 

2.  The  coefficient  of  friction  is  independent  of  the  normal 
pressure  between  the  two  surfaces  and  of  the  extent  of  the 
contact.  This  law  is  not  exactly  true;  especially  for  such  low 
pressures  at  which  a  considerable  part  of  the  sliding  resistance 
is  due  to  adhesion,  and  for  high  pressures  which  result  in  a 
change  in  the  character  of  the  surfaces;  also  when  lubrication 
is  excessive,  for  then  the  friction  is  mixed,  being  neither  "solid" 
nor ''fluid." 

157.  Determination  and  Values  of  the  Coefficient. — The 
coefficient  of  friction  for  two  surfaces  may  be  determined  by 
measuring  their  angle  of  repose  (see  fig.  126).  The  tangent  of 
the  angle  is  the  coefficient  sought.  Or,  the  two  bodies  may 
be  placed  as  in  fig.  125,  and  then  measuring  the  pull  P  neces- 
sary to  start  the  block,  F'  is  known  since  F'  equals  that  pull. 
Also,  N  =  W',  hence  f  =  P/W. 

In  one  of  these  two  ways  many  determinations  have  been 
made,  the  values  of  the  coefficient  for  a  few  materials  ranging 
as  follows: 

Wood  on  wood,  soaped 0.22-0. 44 

"■       "       "       dry 0.30-0.70 

Metal  on  metal,  dry 0.15-0.24 

"       "        "        as  in  polished  and  well- lubricated 

bearings o .  05  -  o .  08 

Wood  on  metal,  dry o .  60 

Hemp  rope  on  wood o .  50  -  o .  80 

Sole-leather  on  wood  or  cast  iron,  as  in  packings,  dry  o .  40  -  o .  60 

Leather  belting  on  pulleys 0.25-1. 00 

Stone  on  stone,  as  in  arches o .  40  -  o .  60 


152     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VL 

EXAMPLES. 

1.  If  the  block  represented  in  fig •12 7  weighs  100  lbs.,  6  is 
p    10°,  and  the   coefficient   of  friction  is  0.2,  how 

great  must  P  be  to  start  the  block? 

Solution:  Suppose  motion  to  be  impending, 

then  F'  =  o.2N\  and  since  the  block  is  at  rest, 
P  cos  io°  — F'  =  o  and  P  sin  io°+7V— 100  =  0.  Solution  of  these 
three  equations  gives  P=  19.62  lbs. 

A  force  slightly  greater  than  this  will  start  the  block. 

2.  Solve  ex.  i  if  the  sense  of  P  is  reversed. 

Ans.  21.05  lt)S. 

3.  If  Pin  ex.  2  is  15  lbs.,  how  great  is  the  friction? 

4.  What  is  the  least  value  of  P  acting  as  shown  in  fig.  127 
that  will  start  the  block?  What  is  the  corresponding  value 
of  l9?  Ans.  19  =  11°  i8'.6. 

5.  If   the  block  represented  in  fig.   128   weighs  -fp 
100  lbs.,  /?  =  25°,/  =  J,  and  ^  =  0,  how  great  must  P 
be  to  start  the  block?                 Ans.  72.46+  lbs. 

6.  To  prevent  slipping,  how  great  must  P  be? 

7.  If  Pis  20  lbs.,  determine  the  friction.  ^Z.     "     T" 

*  r  IG.    125. 

Ans.  22.26  lbs. 

8.  If  P  is  50  lbs.,  determine  the  friction. 

9.  Take  6=10°  in  ex.  5,  and  solve.  Ans.  69.51  lbs. 

10.  Show  that  P  (fig.  128)  to  start  the  body  up  is  a  mini- 
mum if  d  =tan~^/. 

11.  Two  bodies  connected  by  a  cord  are  placed  upon  a 
plane  inclined  12°  to  the  horizontal,  the  string  being  taut  but 
without  initial  tension  and  inclined  12°  with  the  level  base. 
If  the  bodies  weigh  10  and  15  lbs.  and  the  corresponding  coeffi- 
cients of  friction  are  \  and  J,"  determine  the  frictions. 

12.  In  the  preceding  example,  change  the  coefficient  \  to 
\  and  suppose  the  lower  body  to  be  the  lighter  one.  Solve  and 
also  determine  the  tension. 

13.  A  bar  weighing  100  lbs.  rests  upon  two  end  supports  at 
the  same  level.  Suppose  a  force  is  applied  to  it  so  that  the 
action  line  passes  through  the  points  of  support.  If  the  coeffi- 
cients of  friction  for  the  rubbing  surfaces  be  0.2  and  0.25,  how 


§  VIII.] 


ROUGH  SUPPORTS;    FRICTION. 


153 


great  must  the  force  be  to  move  the  bar?  What  can  you  say 
about  the  frictions  when  the  force  is  18  lbs.  and  when  it  is 
20  lbs.? 

14.  How  great  must  P   (fig.    129)   be  to  start  the  wedge 
against  the  force  Q? 


*i-r(a) 

Fig.   129. 

Solution:  When  the  wedge  is  about  to  slip,  the  directions 
of  the  resistances  at  the  rubbing  surfaces  are  known,  for  the 
inclination  of  each  to  the  normal  to  the  surface  on  which  it  acts 
equals  the  corresponding  angle  of  friction.  Of  the  three  forces, 
Q,  R' ,  and  R'\  applied  upon  the  upper  body,  the  direction  of  all 
and  the  magnitude  of  one,  Q,  are  known.  Their  force  triangle 
determines  the  magnitude  and  sense  of  R'  and  i?"  (see  "fig.  1296). 
Having  determined  R" ,  P  and  R'"  may  be  determined  by  means 
of  the  force  triangle  for  those  forces  (see  fig,  129c). 

15.  How  great  must  P  be  to  prevent  the  wedge  from 
slipping  out,  the  wedge  angle  being  25°,  and  all  <^'s  10°  ? 

16.  Show  that  the  wedge  will  not  slip  out 
if  its  angle  is  less  than  (j>"  -\-cj)'"  when  P  =  o. 

1 7 .  Fig.  130  represents  a  jack-screw.  How 
great  a  couple  whose  plane  is  horizontal  must 
be  applied  to  the  screw  to  "  overcome  "  Q? 

Solution:  At  each  point  of  the  lower  sur- 
faces of  the  thread  on  the  screw,  the  nut 
exerts  a  pressure  whose  normal  and  tan- 
gential components  call  dN  and  dF  respect- 
ively. When  the  tendency  of  the  screw  is  ^^^'  ^^°* 
to  rise,  dF  acts  downward.     Let  C  denote  the  moment  of  the 


f54    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 


couple,  p  the  pitch  angle,  and  r  the  average  arm  of  the  frictions 
and  normal  pressures  with  respect  Xo  the  screw  axis.  From 
the  conditions  of  equilibrium  (see  art.  125), 

IFy  =  -Q-I{dF  ^inp)  +  I{dN  cos>P)  =  o; 
IMy  =  C-I{dF  cos  p-r)  - 1 {dN  sin  p-r)=o] 

and,  when  slipping  is  impending,  dF  =  fdN.  .    - 

These  three  equations  make 

C  =  Qr(sin  /?  +  /  cos  ./?)/(cos  /?-/  sin  /?). 

18.  Show  that  to  cause  the  screw  to  descend, 

C  =  Qr{f  cos  /?-sin  /?)/(/  sin  /?  +  cos  /?). 

19.  Show  that  if  P>(j),  the  screw  will  descend  under  the 
action  of  Q  alone. 

20.  Fig.  131  represents  a  lever  supported  in  a  triangular 
bearing.     How  great  a  force,  P,  is  required  to  ''  overcome"  Q? 

Solution:  When  slipping  is  about  to  occur,  the  reactions 
at  A  and  B  act  in  the  directions  indicated.  Since  the  action 
lines  of  P,  Q»  -^'»  ^^^  F"  and  the  magnitude  of  Q  are  known, 
the  magnitude  of  P  (and  of  R'  and  R")  may  be  determined  (see 
art.  141)- 


Fig.  131. 


CD=r 

CB=rsin(j> 

Fig.   132. 


158.  Friction  Circle. — Fig.  132  represents  a  journal  and  its 
bearing  (also  a  pin  joint),  the  fit  being  loose  so  that  the  con- 
tact is  along  a  line  practically.  Let  r  denote  the  radius  of  the 
journal  and  <j)  the  angle  of  friction  for  the  rubbing  surfaces. 
Then  a  circle  concentric  with  a  cross-section  of  the  journal 
whose  radius  equals  r  sin  (j>  is  called  the  friction  circle  for  the 
journal  and  bearing.  It  is  useful  in  solving  certain  problems 
involving  the  friction  of  a  loose-fitting  journal  and  bearing,  pin 
joint,  etc. 


§VIIL]  ROUGH  SUPPORTS;    FRICTION.  155 

Proposition. — When  slipping  is  about  to  occur  at  a  loose 
bearing,  the  action  line  of  the  resistance  offered  by  the  bearing 
is  tangent  to  the  friction  circle. 

•  Proof:  The  resistance  is  applied  at  the  line  of  contact,  rep- 
resented by  A,  and  it  makes  an  angle  cj)  with  the  normal  AC 
(art.   154).     If  AB  is  tangent  to  the  circle,  then 

sin  B AC  =  r  sin  (f)/r  =  sin  ^,     or     BAC  =  cf); 

hence  the  tangent  line  and  the  action  line  of  the  resistance 
coincide. 

Remembering  that  ^he  tangential  component  of  the  re- 
sistance (the  friction)  opposes  the  tendency  to  slip,  the  student 
will  have  no  difficulty  to  tell  which  one  of  the  two  tangent  lines 
which  may  be  drawn  is  the  action  line  of  the  resistance. 

EXAMPLES. 

1.  Fig.  133  represents  a  lever  supported  in  a  loose  cylindri- 
cal bearing.     How  great  a  force  --Vc 

Pis  required  to  ''overcome"  Q?  /     \  Z^^^''^ 

Solution:    There   being   three      >>^^^^^^/^^\X.-^:x^^ 

forces  applied  to  the  lever,  P,Q,  ^^      ^^    ,''    V  L^ 

and  the  resistance  of  the  bearing  ^^^^k^~'\M 

R,  their  action   lines  intersect  in  ^^^P^ 

a  point  (art.  125);  hence  7^  passes  ^^"  ^^^' 

through  D.     Since  R   is   also  tangent  to  the  friction  circle,  its 

action  line  is  determined,  and  the  system  P,  Q,  R  may  be  solved 

for  the  unknown  magnitudes. 

2.  Let  MCN  (fig.  133)  equal  90°,  CM  2  ft.,  CN  6  in.,  the 
angles  CND  and  CMD  60°  and  90°  respectively,  radius  of 
axle  2  in.,  the  coefficient  of  friction  h,  and  Q  1000  lbs.  Deter- 
mine P  and  the  friction. 

3.  Determine  the  largest  force  P  which  Q  will  overcome, 
data  as  in  ex.  2. 

4.  When  the  lever  is  straight  (iVCM=i8o°)  and  P  and  Q 
act  at  right  angles  to  NCM,  show  that 

P(CM Tr  sin  cj))=Q(CN±r  sin  (l>), 

according  as  the  impending  motion  is  with  P  or  Q. 


156    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM,    [Ch.  VI. 


Fig.  134. 


159.  Cone  of  Friction. — Let  P  (fig.   134)  denote  the  result- 
B  A      a-nt   of   all  the   forces  ©applied  to  the  body  repre- 

sented, not   including  the  resistance  of  the  sup- 
porting surface.       Then  the   cone  whose   apex  is 
at  C,  whose   axis   is   the   normal  through  C,  and 
whose  apex  angle  equals  twice  the  angle  of  friction 
is  called  a  cone  of  friction.     In  the  figure  the  cone 
is  represented  by  ACB. 
Proposition. — If  the  action  line  of  the  resultant  of  all  the 
forces  applied  to  a  body  not   including  the  resistance  of  the 
support  falls  within  the  cone  of  friction,  sliding  will  not  occur; 
if  without,  it  will  occur. 

Proof:  The  force  causing  or  tending  to  cause  sliding  is  the 
horizontal  component  of  P  (fig.  134),  equal  to  P  sin  d.     Since 
the  maximum  sliding  resistance,  or  friction,  is 
F'=tan  ^.A/"  =  tan  <j)'P  cos  6, 
(sliding  force)/F'  =  tan  ^/tan  0.     Therefore 
a  d>^  (P  falls  without  the  cone), 

the  sliding  force > limiting  friction; 
a  d<i(j)  (P  falls  within  the  cone), 

the  sliding  force  <  limiting  friction. 

EXAMPLES. 

I.  In  fig.  128  suppose  that  d  and  ^  =  30°,  P  =  5o  lbs.,/  =  J, 
and  the  body  on  the  incline  weighs  100  lbs.  Determine  whether 
the  body  will  slide  and  the  value  of  the  friction  graphically. 

2.  Let  P  in  the  preceding  example  be  10  lbs.,  and  solve. 

3.  A  prismatic  block  of  wood  is  sawed  into  two  parts  so  that 
the  cut  is  inclined  at  an  angle  6  with  the  ends.  If  the  two  parts 
are  laid  together  matching  and  end  pushes  are  appHed  along 
the  axis,  for  which  values  of"^  will  slipping  not  occur? 

4.  Fig.  135  represents  a  slider  which  may  slide  in  the  guides 
A  and  B;    cjy'  and  ^"  are  the  angles  of 
friction  for  the  rubbing  surfaces  respect- 
ively.    Show  that  any  horizontal  force 
applied  above  C  cannot  move  the  slider. 

Solution:  Imagine  slipping  about  to 
occur  at  B;  then  the  resistance  there 
acts   along  the  line  BC.     To    keep  the 


Fig.  135. 


§  VIII.] 


ROUGH  SUPPORTS;  FRICTION. 


157 


slider  at  rest,  the  resistance  at  A  must  act  through  y,  and  since 
the  line  Ab'  is  within  the  cone  at  A,  such  resistance  is  possible, 
and  equilibrium  will  be  preserved. 

5.  A  boy  climbs  a  ladder  which  rests  on  a  horizontal  floor 
and  against  a  vertical  wall.  Determine  graphically  how  far 
he  can  ascend  without  causing  the  ladder  to  slip.  Suppose 
that  the  length  of  the  ladder  is  20  ft.,  its  centre  of  gravity  is 
8  ft.  from  the  foot,  its  inclination  is  45°,  its  weight  and  that  of 
the  boy  are  equal,  and  the  coefficients  of  friction  for  the  sur- 
faces at  the  top  and  bottom  are  0.4  and  0.6  respectively. 

Ans.  About  19  ft. 

6.  A  uniform  ladder  rests  with  one  end  against  a  rough 
horizontal,  the  other  against  an  equally  rough  vertical  plane. 
Determine  the  least  coefficient  of  friction  that  will  allow  the 
ladder  to  rest  in  all  positions. 

Ans.  I. 

7.  Fig.  136  represents  a  bar  AB  resting  in  a  horizontal 
position  upon  two  inclined  planes, 
0'  and  (j>"  being  the  angles  of  fric- 
tion at  A  and  B  respectively.  Show 
that  if  the  weight  of  the  bar  is  neg- 
lected, any  body  suspended  from  a 
point  between  m  and  n  will  not  cause 
it  to  slide,  but  that  if  suspended 
beyond  m  ov  nit  will  cause  sliding. 


1 


dF, 


\  P^^  I 


y 


137- 


8.  The  front  of  a  drawer  is  4  ft., 
_j:  and  its  sides  10  in.  long.  If  the 
angle  of  friction  for  the  rubbing 
siy-faces  at  the  sides  is  22°,  and  the 
drawer  handles  are  3  ft.  apart,  show 
that  the  drawer  cannot  be  opened 
by  a  forward  pull  at  one  handle. 

160.  Belt  Friction. — Fig.  137(a) 
represents  a  cylinder  about  a  part 
of  which  a  belt  (or  cord)  is  wrapped. 
If  the  cylinder  is  not  smooth,  the 
pulls  Ti  and  T^  may  be  quite  Un- 
equal without  causing  slipping  of 


the  belt,  as  may  easily  be  verified  by  trial. 


158    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

The  forces  acting  upon  the  part  of  the  belt  in  contact  with 
the  cyhnder  consist  of  the  tensions  T^  and  T^,  the  normal 
pressure  and  the  friction  (see  fig.  1376).  Let  p  denote  the  nor- 
mal pressure  per  unit  length  of  arc ;  then  the  normal  pressure 
on  any  part  whose  length  is  ds  (fig.  137c)  is  pds.  The  friction 
on  that  part  may  be  called  dF  and  the  tensions  T  and  T  +  dT. 
Since  the  portion  is  in  equilibrium, 

pds  =  2T  sin  —  =  Tdd ; 

hence  p  =  T/r; (i) 

that  is,  the  normal  pressure  per  unit  length  at  any  point  of  the 
contact  equals  the  belt  tension  there  divided  by  the  radius  of 
the  cylinder. 

When  slipping  is  impending,  dF  =  f-pds,  and  since  dF  =  dT, 

dT==f-ds,     or     %  =  i~=fdd. 
'  r  1  r      ' 

Integration  gives  log,,  T      ""  =  j\  0  \  (see  fig.  137a); 

hence  log^  r2-log^  Ti=//?, (2) 

or  .  T,=  T,ef^ (3) 

The  angle  ^  must  be  expressed  in  radians;  e  is  the  base  of  the 
Naperian  system  of  logarithms,  2.718.  The  formulas  apply 
also  when  /?  is  greater  than  271,  that  is,  when  the  cord  more  than 
encircles  the  cylinder. 

For  a  given  value  of  T^,  T^  increases  very  rapidly  with  ^ 
as  shown  by  fig.  138  which  represents  the  locus  of  equation  (3). 
T^  and  /?  are  the  variables  and  e,  f,  and 
T^  constants,  /  being  taken  as  J  and 
T^  =  OA.  To  the  scale  0^  =  7^,  OB  rep- 
resents T2  when  j3  =  A0B. 


EXAMPLES. 

I.  Compute  the  ratio  between  T^  and 
7^2,  when  /  is  ^  and  the  cord  is  wrapped 
twice  around  the  cylinder. 
2.  Plot  in  fig.  138  the  locus  of  equation  (3)  when  /  is  J 


§IX.] 


FORCES  IN  SPACE  AND  MISCELLANEOUS. 


159 


§  IX.    Forces  in  Space  and  Miscellaneous. 

161.  Examples  Involving  Non-Parallel  Non-Coplanar  Forces. 

— The  principles  for  solving  such  examples  are  stated  in  art.  125. 
Sometimes  in  simple  cases  the  example  can  be  resolved  into 
others  the  forces  in  which  are  coplanar.  Such  separation  is 
usually  a  simplification;   ex.  3  is  an  illustration. 

EXAMPLES. 

I  I .  Determine  the  relation  between  P  and  W  of  the  windlass 
represented  in  fig.  139  and  the  reactions  of  the  bearings  in  terms 


Fig.  139. 

of  P.     Take  P  always  at  right  angles  to  the  crank  as  shown, 
and  neglect  friction. 

Solution:  The  forces  acting  on  the  windlass  are  P,  W,  and 
the  reactions  of  the  bearings,  A  and  B\  the  weight  of  the  wind- 
lass is  neglected.  Let  R'  and  R^'  denote  the  reactions  at 
A  and  B  respectively,  and  a  the  angle  which  the  crank  makes 
vith  the  X  axis;   then 

i'F^=  -Psina:+i^:^+i?x"  =  o, 

IFy=  P  cos  a  +Rl-\-Rl'-'W  =  o, 
IM^=  -Pcosa'b-Wx+Ria  =  o, 
IMy=  -Psma'b-RJa  =  o, 

IM,=     Pc-Wr=o, 


r  being  the  distance  from  the  axis  of  the  rope  to  that  of  the 
windlass. 

The  2-resolution  equation  vanishes  since  none  of  the  force? 
have  z  components.     Solving  the  equations,  we  find  that 


i6o    APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 


W  =  Pc/r, 
R^=  -P  sin  a-h/a,  R^  =P{b  cos  a  +cx/r)/a, 

R%^     Psma(i+6/a),  Ri'  =  P(i~x/a)c/r-P  cosa(i+b/a). 

2.  Fig.  140(a)  represents  a  derrick  consisting  of  a  post  (AB), 
boom  (AC),  two  "stiff  legs"  (BD  and  BE),  and  hoisting  cables 
which  are  not  shown  in  detail.  Determine  the  forces  on  the 
parts  of  the  derrick  due  to  the  load  W,  assuming  for  simplicity 
that  B  and  C  are  merely  connected  by  a  cable,  that  the  load  is 


R  COS  AS' 


Fig.  140. 

suspended  from  C,  and  that  the  boom  is  connected  to  the  post 
at  its  lower  end  practically. 

Solution:  The  external  forces  on  the  whcie  derrick  are  W 
and  the  reactions  at  A,  D,  and  E.  The  directions  of  these  reac- 
tions will  depend  on  the  nature  of  the  supports  at  A,  D,  and  E', 
we  will  assume  that  these  are  such  that  the  reaction  at  A  is  d 
single  force  acting  through  A,  and  that  those  at  D  and  E  act 
along  DB  and  EB  respectively. 

We  first  resolve  this  system  into  two  component  systems, 
one  coplanar  (its  plane  being  that  of  the  ground)  and  one  par- 
allel (its  forces  being  vertical).  Let  the  reactions  at  D  and  E 
be  called  R^  and  i?"  respectively;  then  the  components  of  the 
R'  are 


R' 
R' 


cos  45°  acting  in  the  line  AD, 
sin  45°      "      vertically  at  J9, 

and  the  components  of  R"  are 

i?"  cos  45°  acting  in  the  line  >IE, 


R"  sin  45° 


verticallv  at  E. 


§ix.] 


FORCES  IN  SPACE  AND  MISCELLANEOUS. 


ibi 


Let  the  reaction  at  A  be  called  R"\  and  imagine  it  resolved  into 
X,  jy  and  z  components  at  A .  The  coplanar  component  system 
consists  of  forces  as  shown  in  fig.  140(6),  and  the  parallel  com- 
ponent system  consists  of  four  vertical  forces,  R'  sin  45°  at  D, 
R"  sin  45°  at  E,  R^/'  at  A,  and  W  at  C  (all  not  shown). 

'Each  of  the  component  systems  being  in  equilibrium,  we 
may  write  the  appropriate  equations  of  equilibrium  for  each; 
thus  for  the  first  (concurrent) 

IF^=  -R'  cos  45°  cos  45°-^"  cos  45°  cos  45°+i?x'"  =  o, 
IF,=  -R'  cos  45°  sin  45°+^"  cos  45°  sin  45°+i?r'.  =  o; 

and  for  the  second 

IFy=Rl"-W-R'  sin  45°-i^"  sin  45°  =  o, 
IMx=Wb  cos  ^  sind-R'  sin  45° -a  cos  45° 

+ 7?"  sin  45°  •  a  cos  45®  =  o, 

2^/,=  -Wb  cos  ^  cosd+R'  sin  45° -a  cos  45° 

+  i^"  sin  45° -a  cos  45^=0. 

These  equations  determine  the  five  unknown  forces  R\  i?", 
i?^",  Ry\  and  R^'\     The  solution  is  left  to  the  student. 

The  tension  in  the  cable  BC  and  the  pressure  between  the 
post  and  the  boom  may  be  determined  from  a  consideration  of 
the  forces  acting  on  the  boom.  As  these  forces  are  coplanar 
their  determination  is  left  to  the  student. 

3.  Fig.  141(a)  represents  a  "shear-legs  crane."  It  consists 
of  two  posts,  CD  and  CE,  hinged  together  at  the  top  and  hinged 


J 

Fig.  141. 

at  their  bases  so  that  they  can  rotate  about  the  line  joining  D 
and  E.  The  stay  AC  may  be  a  cable  or  a  "stiff  leg";  if  stiff, 
AC  is  constant  in  length  and  the  load  is  swung  in  or  out  (the 


i62     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI.  \ 

posts  rotating  about  DE)  by  moving  the  lower  end  of  A  along  the  \ 
track  AB.  Determine  the  forces  oif  the  parts  due  to  the  load  W.l 
Solution:  Imagine. that  the  two  posts  are  replaced  by  a  sin-] 
gle  one  CB,  and  determine  the  tension  in  the  stay  and  the  com- 1 
pression  {R)  in  CB  from  a  consideration  of  the  forces  applied  I 
(see  fig.  1 416).  Then  resolve  this  compression  R  into  two] 
components  whose  action  lines  coincide  with  the  axes  of  thej 
posts  (see  fig.  141c).  \ 

4.  In  fig.  141  take  AC=<,o  feet,  BD  =  BE=io  feet,  <9  =  6o°,  ] 
T^=io  tons,  and  solve  the  preceding  example  graphically.  ; 

5.  Fig.  142  represents  a  small  dipper  dredge,  the  side  eleva- j 
tion  (a)  representing  a  position  for  filling  the  dipper,  and  theJ 


Scale,  ljn«20  ffe 

Fig.  142.  I 

front  elevation  (6)  a  position  for  emptying.  The  boom  swings  ] 
about  a  vertical  axis  at  its  lower  end,  a  ratchet  on  the  lower  f 
side  of  the  dipper  handle  engages  a  pinion  on  the  boom  by  ■ 
means  of  which  the  effective  length  of  the  handle  can  be  changed,  i 
and  the  two  back  stays  BC  are  fastened  at  points  10  feet  apart,  i 
36  feet  to  the  left  of  D.  In  (a)  assume  that  the  dipper  is  stuck,  ; 
the  pull  in  the  chain  being  5000  lbs.,  and  in  (6)  that  the  dipper  \ 
load  is  1000  lbs.  Neglect  the  weights  of  all  parts  and  deter-  ^i 
mine  in  each  case  the  tensions  in  the  stays  AB  and  BC,  the  j 
compressions  in  the  posts  of  the  "A  frame,"  the  reaction  at  j 
the  pivot  at  the  base  of  the  boom,  and  the  pressure  on  the  ; 
ratchet.     Solve  graphically. 

6.  Fig.   143(a)  represents  a  giant  wharf-crane.     The  struc-  \ 
ture  consists  of  a  rigid  framework,  T-shaped,  and  a  rigid  tripod,  i 


§IX.] 


FORCES  IN  SPACE  AND  MISCELLANEOUS. 


163 


The  stem  of  the  T  stands  within  the  tripod  and  rests  against 
the  tripod  head  and  on  rollers  at  the  base.  Determine  the  reac- 
tions at  the  base  of  the  tripod  legs,  as  far  as  possible,  due  to  a 
load  P  of  150  tons  at  the  position  shown  in  the  figure.  When 
the  cross-piece  of  the  T  is  represented  in  plan  by  Aa,  the  short 
arm  being  on  the  side  A ,  what  are  the  reactions  ? 


70^ 


Fig.  143. 

7 .  The  crane  of  the  preceding  example  is  revolved  by  means 
of  a  circular  rack*  on  the  tripod  head  and  two  pinions  at  oppo- 
site ends  of  a  diameter  of  the  rack.  It  is  estimated  that  when 
the  crane  is  being  turned  against  a  maximum  resistance  (due 
to  friction,  inertia,  and  wind-pressure),  the  reaction  on  the  rack 
is  a  horizontal  couple  of  112,000  foot-pounds.  The  diameter 
of  the  rack  being  22  feet  compute  the^reactions  at  the  base  of 
the  tripod  legs  due  to  the  reaction  mentioned. 

162.  Miscellaneous. — The  principles  for  the  solution  of  the 
following  examples  are  given  in  art.  125.  Similar  examples 
have  been  worked  in  the  first  part  of  this  chapter,  and  the  stu- 
dent should  have  no  difficulty  in  solving  the  following  set. 

EXAMPLES. 

I.  Fig.  144(a)  represents  a  platform  scale  which  consists 
o£  a  frame  FF  with  five  knife-edges,  one  at  K^,  two  at  K^,  and 


1 64     APPLICATIONS  OF  THE  PRINCIPLES  OF  EQUILIBRIUM.    [Ch.  VI. 

two  at  Ky  The  platform  rests  on  three  levers  which  bear  upon 
the  four  knife-edges  at  K^  and  K^.  Xhe  short  levers  are  also  sup- 
ported by  the  long  one  at  5,  and  the  long  one  is  connected  by 


|«— e— ^a>|< — 


Fig.  144(a). 


a  vertical  rod  to  the  "scale-beam"  PK^.  Determine  the  rela- 
tion between  the  weight  of  the  poise  (P)  and  that  of  the  body 
(W)  on  the  platform,  and  show  that  it  is  independent  of  the 
position  of  the  body. 

2 .  Assume  that  the  steam-pressure  P  just  balances  the  load 
W  on  the  hoisting-engine  (fig.  144^).  If  P  =  iooo  lbs.  and  the 
angle  BCA  is  60°,  compute  the  compression  in  the  connecting- 
rod,  the  pressure  against  the  cross-head  guide,  the  tangential 
component  of  the  crank-pin  pressure,  and  W. 


wQ  w 

Fig  i44(&).  Fig.  144(c)- 

3.  Fig.  144(0  represents  a  "crab  hook."  The  lengths  AB 
and  EC  are  12  and  21  in.  respectively,  the  angle  ABC  is  100°, 
CD  =  12",  and  BB'  =  ^  ft.     Determine  the  stresses  in  CD  and 


^  §  IX.] 


FORCES  IN  SPACE  AND  MISCELLANEOUS. 


165 


BB'  if  T1^=iooo  lbs.  and  AA'  =  i2>'\  the  weight  of  the  parts 
being  neglected. 

4.  Fig.  i44(c^)  represents  a  simple  elevator-car.     What  are 
the  pressures  of  the  wheels  on  the  rails  due  to  the  load  VF? 


Fig.  i44(^.  Fig.  i44(^). 

5.  Fig.  i44(^)  represents  a  hand-press,  consisting  of  a  lever 
ACP  and  two  'short  links  BC  pinned  to  a  cross-head  at  their 
lower  ends;  AC=^BC=i$".  Compute  the  pressure  on  the  bale 
in  the  press  and  the  pin -pressures  when  AB  =  2  ft.,  and  the  arm 
of  P  with  respect  to  ^  is  5  ft.,  P  being  100  lbs. 


KINEMATICS. 


CHAPTER  VII. 
RECTILINEAR  MOTION  OF  A  PARTICLE. 

§  I.     Velocity  and  Acceleration. 

163.  Specification  of  Position. — The  position  of  a  point  in  a 
given  line  can  be  specified  by  a  single  quantity,  namely,  the 
abscissa  of  the  point  with  respect  to  any  other  point  in  the  line 
assumed  as  "origin."  Thus  if  the  abscissas  of  points  to  the 
right  of  the  origin  O  (fig.  145)  be  given     ^,  , 

the  plus  sign  and  those  of  points  to  the  -" • : ^ 

left  the  minus  sign,  then  P'  is  specified  ^^^-  ^45- 

by  the  abscissa  J  inch  and  P"  by  —  J  inch.     Position  abscissas 
will  be  denoted  by  x. 

164.  Space-Time  Curves. — A  rectilinear  motion  of  a  point 
can  be  well  represented  by  means  of  a  line  called  the  space-time 
curve  for  the  motion.  This  is  a  line  the  ordinate  and  abscissa 
to  any  point  of  which  represent  the  position  abscissa  and  corre- 
sponding  value    of    the   time    respect- 

yj  |\  ively.      To    construct    this    curve   plot 

/  }«?!        [     \     T      corresponding  values  of  x  and  the  time 
tj-^  j        \         along  vertical  and  horizontal   axes   as 

f|—  \      shown   in   fig.    146,   and  join   all   such 

plotted  points;    the  connecting  line  is 
Fig.  146.  \;\^Q  space-time  curve. 

Evidently  the  space-time  curve  for  a  motion  gives  the  posi- 
tion of  the  moving  point  at  each  instant,  and  it  is  therefore  a 
complete  record  of  the  niotion. 

165.  Displacement. — If  x^  denotes  the  abscissa  of  a  moving 
point  at  the  instant  t^,  and  x^  that  at  a  later  instant  t^,  then  the 

167 


1 68 


RECTILINEAR  MOTION  OF  A  PARTICLE.       [Chap.  VIL 


displacement  for  the  interval  t^  —  t^  is  defined  as  x^—x^  Evi- 
dently a  displacement  {x^  —  x^  may*  be  positive  or  negative; 
hence,  two  displacements  must  agree  in  sign  as  well  as  in  mag- 
nitude to  be  equal. 

1 66.  Kinds  of  Rectilinear  Motion. — If  the  displacements  of 
a  point  in  equal  intervals  of  time  (large  or  small)  are  equal,  the 
motion  is  called  uniform;  and  if  the  displacements  are  not  equal, 
the  motion  is  called  non-uniform.  Non-uniform  motions  are 
further  classified  as  explained  in  art.  172. 

The  space-time  curve  for  a  uniform  motion  is  obviously  an 
inclined  straight  line,  and  the  space-time  curve  for  a  non-uni- 
form motion  is  a  curved  line. 

QUESTIONS. 

'  I.  What  is  the  difference  between  the  motions  represented 
in  fig.  147  (a)  and  (6)? 

2.  What  can  you  tell  of  the  motion  whose  space-time  curve 
is  that  in  fig.  i47(0? 

3.  Are  horizontal  and  vertical  space-time  curves  possible? 


167.  Velocity. — The  velocity  of  a  moving  point  is  the  rate 
with  respect  to  time  at  which  it  changes  position,  or  at  which 
its  displacement  occurs.  Still  otherwise  stated,  it  is  the  time- 
rate  of  (change  of)  the  position  abscissa  of  the  moving  point. 

Let  X  and  t  denote  position  abscissa  and  time  respectively; 
then,  as  shown  in  works  on  calculus  (and  in  Appendix  B),  the 
time-rate  of  ^  is  (i:^/(ir,  hence,  if  ^' denotes  velocity, 

v  =  dx/dt.    , (i) 

This  gives  the  value  of  z;  at  any  instant  t\  its  value  at  a  particu- 
lar instant  equals  the  value  of  dx/dt  for  that  instant. 

If  the  motion  is  uniform^  x  changes  uniformly,  and  the  time- 


§  I.]  VELOCITY  AND  ACCELERATION.  169 

rate  of  x  is  Ax/ At,  Ax  denoting  the  displacement  which  occurs 
in  any  interval  At.     Hence 

v  =  ix/Jt, (2) 

and  plainly  the  velocity  is  constant. 

Since  Ax/ At  is  the  average  time-rate  of  the  displacement, 
equation  (2)  gives  also  the  average  velocity  when  applied  to 
non-uniform  motions.  • 

Since  the  space-time  curve  is  the  "locus"  or  "graph"  of  the 
equation  between  x  and  t,  dx/dt  is  the  general  expression  for  the 
slope  or  gradient  of  that  curve.  Hence  the  velocity  correspond- 
ing to  any  point  on  a  space-time  curve  is  represented  by  the 
slope  of  the  curve  at  that  point.  We  say  "is  represented  by" 
instead  of  equals,  because,  while  the  velocity  at  a  certain  instant 
is  definite,  the  slope  depends  on  the  scale  used  in  plotting  the 
space-time  curve.  The  slopes  must  therefore  be  interpreted  by 
scale  or  be  computed  in  a  certain  way,  as  explained  in  ex.  i, 
art.  169. 

168.  Unit  Velocity. — The  expressions  for  velocity  in  eqs.  (i) 
and  (2)  of  the  preceding  article  imply  a  certain  unit  of  velocity, 
namely,  the  velocity  of  a  point  moving  uniformly  and  so  that 
it  describes  unit  distance  in  unit  time.  Specific  units  of  velocity 
are  one  foot-per-second,  one  mile-per-hour,  etc.  There  are  no 
short  names  for  these  units  except  for  the  nautical  mile-per- 
hour,  which  is  called  a  knot. 

The  term  per  is  conveniently  replaced  by  the  solidus,  /;  foot- 
per-second,  mile-per-hour,  etc.,  are  abbreviated  thus:  ft. /sec, 
mi./hr.,  etc.* 

169.  Sign  of  a  Velocity. — The  expressions  dx/dt  and  Ax/ At 
may  be  positive  or  negative;  therefore  v  must  be  regarded  as 
having  the  same  sign  as  that  of  dx/dt  or  Ax/ At  (see  eqs.  (i)  and 
(2),  art.  167).  When  the  point  is  moving  in  the  positive  direc- 
tion, dx/dt  and  Ax/ At  are  positive,  and  when  it  is  moving  in  the 
negative  direction  they  are  negative  (see  fig.  147);  hence 

the  sign  of  the  velocity  of  a  moving  point  at  any  instant  is 
the  same  as  that  of  the  direction  in  which  it  is  then  moving. 

*  For  dimensions  of  a  unit  velocity  see  Appendix  C. 


I70  RECTILINEAR  MOTION  OF  A  PARTICLE.        [Chap.  VIL 

EXAMPLES. 

1.  What  is  the  velocity  when  ^  =  2f6ecs.  in  the  motion  whose 
space-time  curve  is  shown  in  fig.  148? 

Solution :  We  find  first  the  point  P  of  the  curve  correspond- 
ing to  ^  =  2  sees.,  and  then  draw  a  tangent 
yj^  to  the  curve  at  that  point.     Next  we  take 

/   j  any  point  Q  in  the  tangent  line,  and  from 

/       I  it  draw  a  perpendicular  to  the  horizontal 

p>^____^tv^,^        through   P.     Then  we   measure  by   scale 
^\  ^     ^^    the  lines  QR  and  PR,  and  take  their  ratio 

as  measured.      We  find  that  QR  =  s$  ft. 
X scale :iin.=6o ft.        ^nd  PR  =  4  sccs. ;   hence 

Tecale-.lin.-SBec. 

Fig.  148.  ^' =  3  5/4  =  8-1  ft. -per-sec. 

2.  A  point  moves  so  that  x  =  ct^,  c  being  a  constant.  Show 
that  its  velocity  at  any  time  t  is  sct^. 

3.  Let  c  in  the  preceding  ex.  be  lo,  x  being  in  ft.  and  /  in 
sees.  When  /  =  3  sees.,  where  in  its  path  is  the  moving  point, 
and  what  is  its  velocity?  Ans.  '^  =  270  ft./sec. 

4.  A  point  moves  so  that  x  =  ioot,  x  and  t  being  in  ft.  and 
sees,  respectively.     What  is  its  velocity? 

Solution:  Here  x  varies  uniformly;  hence  v  =  Jx/Jt.  From 
the  law  of  the  motion,  J^  =  iooJ^,  or  z;  =  ioo  ft. /sec.  (Can  the 
value  of  V  be  deduced  from  eq.  (i),  art.  167?) 

5.  A  body  falls  in  a  vacuum  according  to  the  relation  x  = 
16. 1/2,  X  and  /  being  in  ft.  and  sees,  respectively.  What  is  the 
formula  for  the  velocity  ? 

6.  Draw  a  space-time  curve  for  the  motion  of  a  falling  body. 

7.  A  point  moves  so  that  x  =  iot  —  t^,  x  and  /  being  in  ft.  and 
sees,  respectively.     What  is  its  velocity  when  t  =  6  sees.? 

8.  A  point  moves  so  that  ^  =  ccos  {kt),  c  and  k  being  con- 
stants. Deduce  an  expression  for  its  velocity  at  any  time  /. 
Also  let  c=2,  k  =  $,  and  x,  t,  and  kt  be  in  ft.,  sees.,  and  radians 
respectively;  compute  the  vel(5city  when  t  =  4  sees, 

9.  A  sprint  of  100  yards  being  accomplished  in  10  sees.,  what 
was  the  sprinter's  average  velocity  in  ft.-per-sec.  ?  In  mi.- 
per-hr.? 


§1.]  yELOClTY  AND  ACCELERATION,  171 

10.  In  a  certain  motion  v  =  ^f,  v  and  t  being  expressed  in 
ft.-per-sec.  and  sees,  respectively.  Determine  the  displacement 
in  the  interval  from  the  second  to  the  fourth  sec. 

Solution:    Since  v  =  ^t^  =  dx/dt, 

dx  =  7,Pdt,     or     x  =  t^  +  C, 
C  being  a  constant  of  integration  whose  value  depends  on  the 
mode  of  reckoning  x  and  t,  not  specified.*     Let  x^  and  x^  denote 
the  values  of  x  when  t  =  2  and  4  sees,  respectively;   then 

x^  =  A^  +  C  =  6^  +  C,     and     x^  =  2''  +  C=^^+C, 
Hence  rc^  - ^2  =  64  —  8  =  56  ft. 

Instead  of  introducing  a  constant  of  integration  we  might 
integrate  between  limits;   thus,  from  dx  =  $t^dt, 

£yx=^sfU^dt,   or    ^4-^2  =  3[y]]  =  56ft. 

11.  In  a  certain  motion  2;  =  3/^ +  4,  v  and  /  being  expressed 
in  ft.-per-sec.  and  sees,  respectively.  If  the  moving  particle  is 
6  ft.  to  the  right  of  the  origin  at  the  instant  from  which  t  is 
reckoned,  determine  the  position  at  any  time  t  and  draw  the 
space-time  curve  for  the  motion. 

170.  Velocity-Time  Curve. — The  way  in  which  the  velocity  of 
a  moving  point  changes  with  respect  to  time  can  be  represented 
graphically  by  a  line  called  the  velocity-time  curve  for  the  motion. 
This  is  a  line  the  ordinate  and  abscissa  of  any  point  of  which 
represent  the  velocity  and  the  corresponding  value  of  the  time 
respectively. 

To  construct  this  curve  plot  corresponding  values  of  velocity 
(v)  and  time   (t)   along  vertical  and   horizontal  axes   respect- 

*  N'ote  on  the  Determination  of  Constants  of  Integration. — The  student 
is  reminded  that  to  determine  a  constant  of  integration  he  has  only  to 
substitute  for  the  variables  in  an  equation  containing  the  constant  any 
simultaneous  values  of  them  and  then  solve  for -the  constant.  Thus  in 
the  case  above,  suppose  it  had  been  stated  that  x  is  measured  from  the 
place  occupied  by  the  moving  particle  at  the  instant  from  which  /  is 
reckoned;  then  when  t  was  zero  x  was  also  zero,  i.e.,  simultaneous  values 
of  X  and  t  ^re  x=o  and  t=o.  These  substituted  in  the  equation  con- 
taining C  make  it 

o=o'4-C;     hence     C  =0. 


172 


RECTILINEAR  MOTION  OF  A  PARTICLE.       [Chap.  VII. 


ively,  as  shown  in  fig.  149,  and  join  all  such  plotted  points. 
The  connecting  line  is  the  velocity-time  curve. 

171.  Velocity  Increment. — If  7;i*denotes  the  velocity  of  a 
point  at  an  instant  t^,  and  v^  that  at  a 
later  instant  t^,  then  v^  —  v^ (not v^—v^) is 
the  velocity  increment  for  the  interval 
/j  — ^1-  Evidently  a  velocity  increment 
maybe  positive  or  negative;  hence,  two 
velocity  increments  must  agree  in  sign 

^^^-  ^49-  as  well  as  in  magnitude  to  be   equal. 

172.  Kinds  of  Non-Uniform  Motion. — A  non-uniform  motion 
whose  velocity  increments  for  equal  intervals  (large  or  small) 
are  equal  is  called  uniformly  varying;  one  whose  velocity  incre- 
ments are  unequal  is  called  non-uniformly  varying. 

Evidently  the  velocity-time  curve  for  a  uniformly  varying 


(6) 
Fig.  150. 

motion  is  a  straight  line  (fig.  150  a  and  b)  and  that  for  a  non- 
uniformly  varying  one  is  a  curved  line  (fig.  150c). 

QUESTIONS. 

1.  What  is  the  difference  in  the  motions  whose  velocity- 
time  curves  are  shown  in  fig.  150  (a)  and  (6)? 

2.  What  can  you  say  of  the  motion  whose  velocity-time 
curve  is  shown  in  fig.  150(c)? 

3.  Are  horizontal  or  vertical  velocity-time  curves  possible? 
173.  Acceleration. — By  acceleration  of  a  moving  point  is 

meant  the  rate  at  which  its  velocity  changes  with  respect  to 
time,  or  simply  the  time-rate  of  (change  of)  its  velocity. 

Let  V  and  t  denote  velocity  and  time  respectively;  then,  as 
shown  in  works  on  calculus  (and  in  Appendix  B),  the  time-rate 
of  V  is  dv/dt\  hence  if  a  denotes  acceleration, 

a=dv/dt-dVdt' (i) 


§L]  VELOCITY  AND  ACCELERATION.  I73 

Equation  (i)  gives  the  value  of  a  at  any  instant  t\  its  value  at 
a  particular  instant  equals  the  value  of  dv/dt  or  dH/df  for  that 
instant. 

//  the  motion  is  uniformly  varying,  v  changes  uniformly  and 
the  time-rate  of  v  is  Jv/Jt,  Jv  denoting  the  velocity  increment 
for  any  interval  Jt.     Hence  in  this  case  * 

a  =  iv/it, .     (2) 

and  plainly  the  acceleration  is  constant. 

Since  Jv/Jt  is  the  average  time-rate  of  the  velocity,  equa- 
tion (2)  gives  also  the  average  acceleration  when  applied  to 
;  non-uniform  motions  non-uniformly  varying. 

Since  the  velocity-time  curve  is  the  locus  or  graph  of  the 
equation  between  v  and  /,  dv/dt  is  the  general  expression  for 
i  the  slope  or  gradient  of  that  curve.  Hence  the  acceleration 
corresponding  to  any  point  on  a  velocity-time  curve  is  repre- 
sented by  the  slope  of  the  curve  at  that  point.  The  slopes 
must  be  interpreted  by  a  scale,  or  be  computed  in  a  certain  way 
as  explained  in  ex.  i,  art.  175. 

174.   Unit  Acceleration. — The  expressions  for  acceleration  in 

eqs.  (i)  and  (2)  art.  173  imply  a  certain  unit  of  acceleration, 

I  namely,  the  acceleration  of  a  point  whose  velocity  varies  uni- 

;  formly  and  so  that  it  changes  by  a  unit  in  each  unit  time. 

Specific  units  of  acceleration  are 

one  knot-per-hour  (one  nautical  mile-per-hour-per-hour) 
one  foot-per-second-per-second,  etc. 
Abbreviating  the  term  per  as  before,  the  above-named  units  are 
written  thus:   knot/hr.  (mi./hr./hr.),  ft. /sec. /sec;  or  still  more 
briefly,  mi./hr^,  ft. /sec. ^  * 

-  175*  Sign  of  an  Acceleration. — The  expressions  dv/dt  and 
Av/At  may  be  positive  or  negative;  therefore  a  must  be  regarded 
as  having  the  same  sign  as  that  of  dv/dt  or  Av/At  (see  eqs.  (i) 
and  (2),  art.  173).  Now  when  the  velocity  increases  alge- 
braically, dv/dt  and  Av/At  are  positive,  and  when  it  decreases 
algebraically,  dv/dt  and  Av/At  are  negative  (see  fig.  150c); 
hence 


*  For  dimensions  of  a  unit  acceleration  see  Appendix  C. 


174  RECTILINEAR  MOTION  OF  A  PARTICLE.       [Ckap.  VII. 

the  sign  of  the  acceleration  of  a  point  at  any  in- 
stant is  positive  or  negative^  according  as  the  ve- 
locity is  then  increasing  or  decreasing  (algebraically). 


EXAMPLES.  *        \ 

I.  What  is  the  acceleration  when  /  =  3  sees,  in  the  motion  \ 
whose  velocity-time  curve  is  represented  in  fig.  151?  ■ 

^  Solution:  First  we  find  the  point  I 

^^.^^'^]      P  on  the  curve  corresponding  to  ^  =  3  '\ 
^^"^^  sees.,  and  then  draw  a  tangent  to  the  i 

^,^.^f\^^<C '      curve  at  that  point.     Next  we  take  \ 

/   I  \^^  any  point  Q  on  the  tangent  and  draw  i 

! \ I      from  it  a  perpendicular  to  the  hori-  \ 

V  scale:  1  in.»i6  ft.  per  sec.         zontal  line  through  p.    Then  we  meas-  ■ 
T  scale:  iin.=8  sec.  urc  by  scale  the  lines  QR  and  PR  and  ^ 

^^^'  ^51-  take  their  ratio  as  so  measured.     We  : 

find  that  QR  =  7  ft.-per-sec.  and  PR  =  8  sees. ;  hence  j 

0  =  7/8  =  0.875  ft. -per-sec.-per-sec.  i 


2.  In  a  certain  experiment  on  "getting  up  speed"  of  electric  ^ 
trains  the  speeds  recorded  at  5 -minute  intervals  were  as  follows :  ' 
o,  19,  30,  35,  38,  and  40.5  mi.-per-hr.  Draw  the  velocity-time  \ 
curve  for  the  motion  and  determine  the  accelerations  at  the  ^ 
beginning  and  end  of  the  period.  \ 

3.  The  motion  of  a  point  being  according  to  the  equation  i 
x  —  ct^^  c  being  a  constant,  show  that  the  acceleration  equals  tct.  \ 

4.  Let  c  in  the  preceding  example  be  10,  ric:  being  in  ft.  and  /  i 
in  sees.     What  is  the  value  of  the  acceleration  when  ^  =  3  sees.?  I 

5.  A  point  moves  so  that  z;  =  50/,  v  and  t  being  in  ft.-per-sec.  j 
and  sees,  respectively.     What  is  its  acceleration? 

Solution:  Since  v  varies  uniformly,  a  =  Av/At.  From  the  j 
law  of  the  motion  Av  =  50 J/;  hence  0  =  50  ft.-per-sec.^  (Can  the  ) 
value  of  a  be  deduced  from  eq.  (i ) ,  art.  1 73  ?)  \ 

6.  A  body  falls  in  a  vacuum  according  to  the  law  x=i6.it^,  \ 
X  and  t  being  in  feet  and  seconds  respectively.  What  is  the  j 
value  of  the  acceleration?  i 


§  L]  VELOCITY  AND  ACCELERATION,  175 

7.  Draw  a  velocity- time  curve  for  a  body  falling  in  vacuum, 
and  compare  with  the  curve  drawn  in  your  solution  of  ex.  6, 
art.  169. 

8.  A  point  moves  so  that  x=i.ot—t'^,  x  and  t  being  in  ft. 
and  sees,  respectively.  What  is  its  acceleration  when  1  =  6 
sees.  ? 

9.  A  point  moves  so  that  x  =  ccos  (kt),  c  and  k  being  con- 
stants. Deduce  an  expression  for  its  acceleration  at  any  time  /. 
Also  let  r=2,  ^  =  3,  and  x,  t,  and  kt  be  in  ft.,  sees.,  and  radians 
respectively;  compute  the  acceleration  when  ^  =  4  sees. 

10.  The  "acceleration  due  to  gravity"  is  about  32.2  ft.-per- 
sec.-per-sec.     Express  the  same  in  yard-minute  units. 

11.  The  law  of  a  motion  is  a=iot  (ft .-sec.  units),  and  the 
velocity  is  10  ft.-per-sec.  when  /  is  4  sees.  Determine  the  veloc- 
ity at  any  instant. 

Solution :  Since  a  =  dv/dt  —  lot^dv^iotdt, 

and  v=ioJt  dt  =  $t^  +  C, 

Now  7;  =  10  and  /  =  4  being  simultaneous  values  of  v  and  /,  they 
satisfy  the  last  equation;   hence 

10  =  5X4^  +  ^,     or     C*=~7o, 

and  v  =  $i^  —  'jo. 

12.  If  the  law  of  a  motion  is  a  =  10/ +  5  (ft. -sec.  units)  and  x, 
V,  and  /  are  simultaneously  zero,  determine  the  values  of  v  and  x 
at  any  time. 

176.  "Acceleration-Time"  and  Other  Curves. — If  simulta- 
neous values  of  the  acceleration  and  the  time  of  a  motion  be 
plotted  on  two  rectangular  axes,  then  all  such  plotted  points 
determine  a  curve  called  the  acceleration-time  curve  for  the 
motion.  Evidently  it  is  a  graphical  representation  of  the  way 
in  which  the  acceleration  varies  with  the  time. 

Other  curves  descriptive  of  a  motion  can  be  drawn.     Thus 

values  of  the    velocity  {v)  and  the  position-abscissa  (x)  of  a 

moving  point  if  plotted  make  a   "  velocity-space  curve,"  and 

values   of   a   and   x  if   plotted   make   an  "acceleration-space 

. curve." 


176  RECTILINEAR  MOTION  OF  A  PARTICLE.       [Chap.  VII^ 


§  II.     Important  Special  Motions. 


177.  Uniform  Motion. — The  velocity  is  constant,  and  there-^ 
fore  the  acceleration  is  zero  and  the  displacement  (^2~'^i)  ^ 
any  interval  (^2  — ^1)  given  by 

V  denoting  the  velocity.  \ 

178.  Uniformly   Accelerated   Motion. — ^The  acceleration   isj 

constant,  and  the  velocity  increment  {v^—v^  in  any  interval 
(^2-^1)  is  given  by 

(^2-^i)=cf(i2-y. (i| 

a  denoting  the  acceleration.  According  to  this  equation  thd 
velocity  varies  uniformly;  hence  the  average  velocity  for  thd 
interval  {t^  —  t^  is  i(^2  +  ^i),  ^^^  "^^^  displacement  {oc^—x^  in  thd 
interval  is  given  by  ,  \ 

(^-^i)=ife  +  ^i)fe-^i) •     (2^ 

Let  Xq,  Vq,  and  0  be  simultaneous  values  of  x,  v,  and  t.     Since     i 
a  =  dv/dt,     dv  =  adt,     or     v==at  +  Cj^.  j 

Substituting  simultaneous  values  of  v  and  /  in  the  last  equation,! 

we  find  that  Ci  =  Vq]   hence  I 

I 
v  =  at+VQ (3)1 

Since  v  =  dx/dt,  dx  =  at  dt+v^dt;    and  hence  j 

x==iat^+VQt  +  C2.  ^  ; 

Substituting  simultaneous  values  of  x  and  /  in  the  last  equation^ 
we  find  that  Cz^Xq;  hence  ^ 

x==iat^+VQt+XQ.    .......     (4)1 

EXAMPLES.  ] 

A  body  moving  near  the  earth  under  the  influence  of  its! 
attraction  would  have  a  constant  acceleration  were  it  not  for^ 
air  resistance.  In  the  following  examples  neglect  this  resist* 
ance  and  denote  the  acceleration  by  g  and  the  distance  (meas-] 


§  IL]  IMPORTANT  SPECIAL  MOTIONS.  ill 

ured  positively  downward)  of  the  moving  body  from  the  start- 
ing-point by  X. 

I.  If  a  body  falls  from  rest,  show  that 

v  =  gi,     x==^gi^,     and     v^  =  2gx. 
I       2.  If  a  body  is  projected  down  with  a  velocity  v^^  show  that 
I  v  =  gt  +  VQ,    x^lgf^  +  v^t,    and    v^  =  2gx+VQ^. 

3.  If  a  body  is  projected  up  with  a  velocity  z'q,  show  that 
[  v  =  gt-VQ,    x  =  }gP-VQt,    and    v^  =  2gx  +  v,\ 

^  179.  Simple  Harmonic  Motion. — This  may  be  defined  as  a 
rectilinear  motion  in  which  the  acceleration  of  the  moving  point 
is  proportional  to  its  distance  from  an  origin  in  the  path  and  is 
always  directed  from  the  point  to  the  origin.  It  may  also  be 
defined  thus :  If  a  point  travels  in  a  circle  describing  equal  dis- 
tances in  all  equal  intervals  of  time,  then  thje  motion  of  the  pro- 
jection of  the  point  on  any  diameter  of  the  circle  is  a  simple  har- 
monic one.  We  will  choose  the  latter  definition  and  show  in  the 
sequel  that  it  is  in  accord  with  the  former. 

Imagine  P  (fig.  152a)  to  start  from  Pq  and  move  uniformly 
in  the  circle  in  the  counter-clockwise  direction.  The  motion  of 
the  projection  of  P  on  O^F  is  harmonic  and  will  now  be  discussed. 

Let  £  denote  the  angle  XO^Pq,  the  lead  (fag  if  negative); 
id+e)       "         "       "     XO^P,  the  pftase  angle  at  timet; 

y      **         "    ordinate  O^ ,  tht  displacement ; 

T      **        **   time  of  one  revolution  of  P,  the  period; 

n      "        "   number  of  revolutions  per  unit  time,  the  fre- 
quency; 

r      "        "   radius  of  the  circle,  the  amplitude; 

V      **        "   velocity  of  the  projection  (F)  of  P; 

t      "        **   time  elapsed  after  starting. 
If  6  is  expressed  in  radians,  it  is  plain  that 

27T 

6'^^t='27rnt  =  wt, 
CD  being  an  abbreviation  for  27r/T  and  2;rw;    oj  may  be  defined 


178 


RECTILINEAR  MOTION  OF  A  PARTICLE.       [Chap.  VIL 


also  as  the  angle  described  by  O^P  per  unit  time.     From  the  hg- 
ure  it  will  be  seen  that 


y  =  r  sin  {cot +  s:), (i) 


and  since  v  =  dy/dt, 


v  =  ojr  cos  {wt  +  e)  =  (jjr  sin  (a>^+e  +  7r/2) 
=  cox  =^  w{r'^  —  y^)^ . 

Since  a  =  dv/dt, 


}•  ■ 


a=  —  co\  sin  {cot -{■£)  =  co^r  sin  {cot 
=  —  co'^y. 


+  £  +  7r)  ) 


(2) 


(3) 


Equation  {i)  is  represented  in  fig.   152(6).     The  curve  is  a  5 

sinusoid,  and   since   the   ordinates   and   abscissas  denote  "dis-  \ 
placement"  and   time    respectively,  it  is  a  displacement-time, 

or  a  space-time  curve.     The  curve  may  be  constructed  as  fol-  \ 

lows:    Having  drawn  the  "circle  of  reference"  (fig.  152a)  and  | 

having  fixed  the  initial  position  of  P,  divide  the  circumference  \ 

into  a  number  of  equal  parts  (16  is  convenient)  beginning  with  i 

pQ,  and  number  that  point  o  and  the  others  successively  in  the  j 

I 

■I 


§11.]  IMPORTANT  SPECIAL  MOTIONS.  179 

direction  of  the  motion  of  P,  i,  2,  etc.,  16  coinciding  with  o. 
Then  on  the  t  axis  (fig.  1526)  lay  off  any  convenient  length  to 
represent  the  period,  divide  it  in  16  equal  parts,  and  erect  ordi- 
nates  at  the  points  of  division.  Number  these  beginning  with 
O'y,  o,  I,  2,  3,  etc.,  the  last  being  16.  Then  project  points  o, 
1 ,  2,  etc.,  of  the  circumference  of  the  circle  upon  the  correspond- 
ing or  din  at  es.  The  curve  through  the  projections  is  the  dis- 
placement-time curve. 

Equations  {2)  are  represented  in  fig.  152  {c)  and  (J),  {c)  rep- 
resenting v  =  cox  and  {d)  the  other  one.  The  equation  v  =  ajx 
shows  that  the  velocity  at  any  displacement  y  equals  co  times 
the  corresponding  value  of  x.  Hence  to  construct  the  curve  in 
{c)  lay  off  O2V  equal  to  O^V  in  (a)  and  make  Vv  equal  to  loO^H^ 
laying  it  off  up  or  down  according  as  O^H  {x)  is  positive  or  nega- 
tive. Repeat  this  construction  for  several  positions  of  P  and 
thus  determine  the  curve. 

The  curve  in  {d)  is  the  velocity-time  curve  for  the  harmonic 
motion  and  may  be  constructed  in  various  ways:  for  instance, 
lay  off  0"i6  to  represent  the  period,  divide  it  into  sixteen  equal 
parts,  and  number  the  points  of  division  beginning  with  0",  o, 
1,2,  etc.  At  these  points  erect  values  of  the  velocity  which  may 
be  obtained  from  {c)  by  obvious  methods. 

It  is  worth  noting  that  the  eq\xait\on  v  =  ojr  sin  {ojt  +  s -\- n / 2) 
is  very  similar  to  y=r  sin  {iijt-\-s),  v,.  cor,  and  e-|-;r/2  in  the  first 
replacing  y,  r,  and  s  in  the  second.  Hence  the  variation  in  v 
is  simply  harmonic  and  can  be  represented  by  a  harmonic  motion 
whose  period,  phase,  and  amplitude  are  respectively  equal  to, 
90°  ahead  of,  and  oj  times  the  period,  phase,  and  amplitude  of 
the  given  motion. . 

,  Since  (b)  and  (d)  are  y-t  and  v-t  curves  respectively  and 
v  =  dy/dt,  any  ordinate  in  the  latter  figure  should  be  equal  (by 
scale)  to  the  slope  of  the  tangent  to  the  curve  in  the  former  at 
the  corresponding  point.  Thus  the  ordinate  Vv  in  (d)  should 
equal  the  slope  of  the  tangent  at  V  in  (b). 

Equations  (j)  are  represented  in  fig.  152  (e)  and  (/),  (e)  rep- 
resenting a=  —co^y  and  (/)  the  other  one.  Equation  a=  —co^y 
shows  that  the  acceleration  is  proportional  to  the  displacement 
(y)  and  that  it  is  always  opposite  to  the  displacement  in  sign,  i.e., 


l8o  RECTILINEAR  MOTION  OF  A  PARTICLE.      [Chap.  VII. 

it  is  always  directed  from  the  moving  point  toward  the  middle 
of  the  path.  Hence  the  two  definitions  of  simple  harmonic 
motion  previously  given  agree.  . 

To  construct  the  curve  in  [e),  lay  off  O^V  equal  to  y  and  make 
Va  equal  to  (j?y,  laying  it  off  down  or  up  according  as  y  is  posi- 
tive or  negative.  Repeat  this  construction  for  a  number  of 
positions  of  P  and  thus  determine  the  line  O^a. 

The  curve  in  (/)  is  the  acceleration-time  curve  for  the  har- 
monic motion  and  may  be  constructed  in  various  ways:  For 
instance,  lay  off  0"'i()  to  represent  the  period,  divide  it  into  i6 
equal  parts,  and  number  the  points  of  division,  beginning  with 
0"'y  o,  I,  2,  etc.  At  these  points  draw  ordinates  to  represent 
the  corresponding  values  of  the  acceleration  which  may  be  ob- 
tained from  {e)  by  obvious  methods. 

The  equation  a  =  aPrsin  {iDt-\-s,-\-7i)  is  the  same  in  form  as 
y  =  r?,m  {cot  +  s),  a,  aPr,  and  (s  +  tt)  in  the  first  replacing  y,  r, 
and  £  in  the  second.  Hence  the  variation  of  a  is  analogous  to 
that  of  y\  in  fact  the  variation  in  a  is  simply  harmonic  and 
it  can  be  represented  by  a  harmonic  motion  whose  period, 
phase,  and  amplitude  are  respectively  equal  to,  t8o°  ahead  of, 
and  0/  times  the  period,  phase ,  and  amplitude  of  the  given  motion. 
Since  {d)  and  (/)  are  v-t  and  a-t  curves  respectively  and 
a  =  dv/dty  any  ordinate  in  the  latter  figure  equals  (by  scale)  the 
slope  of  the  tangent  to  the  curve  in  the  former  at  the  corre- 
sponding point.  Thus  the  ordinate  Va  in  (/)  equals  the  slope 
of  the  tangent  at  v  in  (J). 

180.  Mechanism  for  Producing  a  Simple 
J  L  Harmonic  Motion. — The  mechanism  repre- 

sented in  fig.  153  consists  of  a  crank  and  a 
slotted  slider,  the  pin  of  the  crank  fitting 
the  slot  of  the  slider.  As  the  crank  is 
rotated  the  slider  moves  up  or  down,  and 
plainly  if  the  crank  rotates  uniformly  every 
point  of  the  slider  executes  a  simple  har- 
monic motion.  ' 
„  The  space-time  curve  for  the  motion  of 
the  slotted  slider  can  be  automatically 
drawn  as  follows:    Fasten  a  pencil  to  the  slider  so  that  it  will 


I   ^ 


V 


§"•] 


IMPORTANT  SPECIAL  MOTIONS. 


i8i 


mark  on  a  plane  surface  as  the  slider  moves,  and  then  cause  a 
sheet  of  paper  to  move  uniformly  over  the  surface  in  a  direction 
at  right  angles  to  that  of  the  motion  of  the  pencil;  the  line 
traced  on  the  paper  is  the  space-time  curve.  If  the  mechanism 
for  moving  the  paper  is  connected  with  the  crank-shaft,  then 
the  curve  traced  by  the  pencil  is  a  space-time  curve  for  a  har- 
monic motion  whether  the  mechanism  is  driven  uniformly  or 
not. 

i8i.  Motion  of  the  Piston  of  a  Steam-engine. — Let  OP  (fig. 
154)  represent  the  crank  and  the  connecting-rod  of  a  steam- 
engine  slider-crank  mechanism,  and  suppose 
that  the  crank  turns  uniformly. 

Let  n  denote  number  of  revolutions  of  the 
crank  per  unit  time ; 
c       "       length  of  crank ; 
r      "  '*       "  connecting-rod; 

y       "       the  distance  of  C  from  0 ; 
t       **         "   time  required  to  describe 
the  angle  PoOP. 
Then  d  =  2nnt  =  ojt,  co  being  an  abbreviation 
for  27rw,  and 

y  =  {r^  —  c^  cos^  ajt)i  +  c  sin  cot (i) 


Hence  the  velocity  of  the  piston  is 


dy 

dt' 


ecu  I  cos  ajt  + 


sm  2  cot 


2(r^/c^  —  cos^  cot)i 


and  its  acceleration  is 
d^y 


df 


=  —CCO'' 


I  sin  Cb 


/  + 


cos^  (xjt  —  {r^ I c^')  cos  2(xji 

(7-2/^2 -COS^O^O^ 


(2) 


(3) 


Close  approximate  values  of  the  velocity  and  acceleration  can 
be  found  as  follows :   Since  approximately 

^(i-pcos^o^^j   =r^i_l_cos2c^^j, 


^=  ('--?.)- 


4r 


sin  (2a>/  +  ;r/2)  +^  sin  wt^  very  nearly.      (4) 


1 82  RECTILINEAR  MOTION  OF  A  PARTICLE.       [Chap.  VII. 

Hence,  approximately,  the  velocity  of  the  piston  is 

dy 

~  =  ca^cos  ojt  —  {c/2r)  Q,os{2(ijt-\-7:/2)\^  ,     .     .     (5) 

and  the  acceleration  is 

--7^= —cco^  [sin  wt  —  (^c/r)  sin  {2a)t-\- 71/2)],     ,     .     (6) 

If  the  connecting-rod  were  infinitely  long,  c/r  would  be  zero 
and  the  second  term  of  eq.  (3)  would  vanish;  hence  the  motion 
of  the  piston  would  be  simply  harmonic.  The  smaller  the  ratio 
c/r  the  more  nearly  is  the  motion  of  an  actual  mechanism  sim- 
ply harmonic. 

EXAMPLE. 

Take  r/c  equal  to  four  and  compute  the  values  of  the  accel- 
eration in  terms  of  co?  from  equations  (3)  and  (6),  when  d  =  o°, 
±30°,  ±60°,  and  ±90°  (see  fig.  154).  Plot  these  values  on  the 
same  base,  draw  the  a-6  curves  and  compare.  Draw  also  on 
the  same  base  the  curve  a=  —  cap'  sin  ix)t. 


CHAPTER  VIII. 
CURVILINEAR  MOTION. 

§  I.     Velocity  and  Acceleration. 

182.  Specification  of  Position. — It  is  usually  convenient  here- 
in to  specify  the  position  of  a  point  in  space  by  Cartesian  coor- 
dinates, but  for  the  purpose  of  this  section  it  is  more  convenient 
to  specify  position  by  means  of  a  vector.  The  vector  drawn 
from  a  fixed  origin  of  reference  to  the  point  to  be  located  is  called 
the  position-vector  of  the  point.  For,  if  the  vector  is  known  the 
position  of  the  point  with  reference 
to  the  origin  is  also  known.  Thus, 
the  direction  of  the  vector  OP  (fig. 
155)  fixes  the  direction  of  P  from  the 
origin  O,  the  length  of  the  vector  fixes 
the  distance  of  P  from  O,  and  thus  the  ^  ^ 
position  of  P  is  determined. 

The  position  of  a  point  in  a  given  ^^^-  ^55- 

line  can  be  specified  by  a  single  quantity :  Thus  if  A  is  an  origin 
of  reference  in  the  line  AP,  the  ab^issa  s  measured  from  A  along 
the  line  fixes  P,  it  being  understood  that  5  is  positive  for  points 
on  one  side  of  A  and  negative  for  those  on  the  other. 

183.  Space-Time  Curve. — A  curvilinear  motion  can  be  rep- 
resented in  part  by  means  of  a  line,  the  space-time  curve  for  the 
motion.  The  ordinate  and  abscissa  of  each  point  of  it  repre- 
sent respectively  simultaneous  values  of  the  position  abscissa  5 
of  the  moving  point  and  of  the  time  t.  It  is  analogous  to  the 
space-time  curve  for  a  rectilinear  motion  (art.  164)  and  is  simi- 
larly constructed. 

This  curve  must  not  be  confounded  with  the  path;  the  latter 
may  be  tortuous,  but  the  former  is  always  a  plane  curve. 

184.  Displacement. — The  displacement  of  a  point  during  an 
interval  in  which  it  moves  from  A  to  B  is  defined  to  be  the  vec- 

183 


1 84  CURVILINEAR  MOTION.  [Chap.  VIII. 

tor  AB,  The  term  therefore  does  not  refer  to  the  path  actually 
described  between  A  and  B,  but  simply  to  the  straight  path 
between  A  and  B.  Observe  that  if  O  is  any  point  of  reference, 
vectorially 

AB  =  OB-OA. 

Two  displacements  must  agree  in  direction  as  well  as  in  magni- 
tude in  order  to  be  equal. 

185.  Kinds  of  Motion. — The  definitions  under  this  title  given 
in  art.  166  refer  to  rectilinear  motion ;  those  following  are  general. 

If  the  displacements  of  a  moving  point  in  equal  intervals  of 
time  (large  or  small)  are  equal,  its  motion  is  called  uniform;  and 
if  unequal,  the  motion  is  called  non-uniform. 

In  order  that  all  displacements  may  be  equal,  the  path  must 
be  straight;  hence  a  uniform  motion  as  here  defined  must  be 
rectilinear. 

186.  Velocity. — By  velocity  of  a  point  is  meant  the  rate  with 
respect  to  time  at  which  it  changes  position  or  at  which  its  dis- 
placement occurs.  Still  otherwise  stated,  velocity  is  the  time- 
rate  (of  change)  of  its  position- vector. 

Let  P(fig.  1 56)  be  a  point  moving  in  the  path  APB,  and  O 
and  0'  points  of  reference ;  then  OP  is  the 
position- vector  of  P,  and  5  its  position-ab- 
scissa. Now  in  general  the  rate  of  the  vec- 
tor OP,  i.e.,  the  velocity,  changes  with  the 
time  as  shown  in  Appendix  B ,  and  when  the 
moving  point  is  at  P  the  rate  is  a  vector 

whose  direction  is  that  of  the  tangent  at  P, 
whose  magnitude  equals    the    time-rate  of 
(change  of)  s. 
If  the  magnitude  of  the  velocity*  be  denoted  by  v,  then 

v  =  ds/dt;       ^  .     .     (i) 

and  if  s  varies  uniformly,  ds/dt  is  constant  and 

v  =  is/Jt (2) 

*  The  word  "speed"  has  been  set  apart  by  several  recent  writers  to 
denote  magnitude  of  a  velocity.  Such  usage  is  very  convenient  and  will 
be  followed  herein. 


§1.] 


VELOCITY  AND  ACCELERATION. 


185 


Observe  that  our  definitions  lead  to  the  result  that  the  veloC' 
ity  in  a  curvilinear  motion  cannot  be  constant,  for  although  its 
magnitude  might  be  constant  its  direction  continually  changes. 
This  result  is  in  accord  with  the  definitions  of  uniform  and  non- 
uniform motion  (art.  185). 

EXAMPLE. 

The  point  P  (fig.  157)  describes  the  circle  in  such  a  way  that 
s  =  2t'^,  s  and  t  being  in  ft.  and  sees,  respect- 
ively, (a)  Deduce  an  expression  for  the 
speed  at  any  time  t.  (6)  What  are  the 
magnitude  and  direction  of  the  velocity 
when  t  =  2  sees.? 

Solution:  (a)  Since  the  speed  {v) 
equals  ds/dt,  from  the  equation  of  the 
motion,  v  =  6t'^.  Fig.  157. 

187.  Speed-Time  Curve  and  Hodograph. — The  way  in  which 
the  speed  of  a  moving  point  varies  can  be  represented  by  a  line 
called  the  speed-time  curve  for  the  motion.  The  ordinate  and 
abscissa  of  each  point  of  it  represent  simultaneous  values  of  the 
speed  and  the  time  respectively.  It  is  analogous  to  the  veloc- 
ity-time curve  for  a  rectilinear  motion  (art.  170)  and  is  similarly 
constructed. 

If  from  any  point  vectors  be  drawn  which  represent  the 
velocities  of  a  moving  point  and  the  free  ends  of  all  such  vectors 


be  joined,  the  joining  line  is  called  the  hodograph  for  the  motion. 
Thus  let  Pi,  P4  (fig.  158)  be  the  path  of  a  moving  point,  and  let 
its  velocities  at  the  points  Pj,  Pj,  etc.,  be  as    represented  by 


^S6  *    CURyiLINEyiR  MOTION.  [Chap.  VIII. 

Pi^i,  ^2^2»  etc.     The  hodograph  is  P,!P^',  O'P^  being  equal  and 
parallel  to  P^v^,  O'P^  parallel  and  equ^l  to  P^v^,  etc. 

A  hodograph  may  be  plane  or  tortuous,  but  a  speed-time 
curve  is  always  plane.  The  latter  curve  corresponding  to  fig. 
158  (a)  and  (6)  is  shown  at  (c). 

EXAMPLES. 

1.  Suppose  that  P  (fig.  157)  describes  the  circle  with  a  con- 
stant speed  of  10  ft.-per-sec.  Draw  a  hodograph  and  a  speed- 
time  curve  for  the  motion  from  A  to  B. 

2.  Draw  a  hodograph  and  a  speed- time  curve  for  a  motion 
from  A  to  B  (fig.  157)  according  to  the  law  stated  in  the  ex.  of 
art.  186,  determining  at  least  four  points  on  the  hodograph. 

188.  Acceleration. — By  acceleration  of  a  point  is  meant  the 
time-rate  (of  change)  of  its  velocity. 

In  deducing  the  expression  for  this  rate  it  must  be  remem- 
bered that  velocity  is  a  vector 
quantity.  According  to  Appen- 
dix B  the  rate  is  a  vector,  and  if 
A'P'  (fig.  159)  is  a  hodograph 
and  O'P'  represents  the  velocity 
at  any  instant  t,  then  the  accel- 
^^^-  ^  59-  eration  at  that  instant  is  a  vector 

whose  direction  is  that  of  the  tangent  to  the  hodograph  at  P', 
whose  magnitude  is  the  time-rate  of  5', 
5'  being  the  distance  of  P'  from  any  fixed  point  on  the  hodo- 
graph as  origin.* 

If  a  be  used  to  denote  the  magnitude  of  the  acceleration, 

a  =  dsVdt; (i) 

and  if  a  varies  uniformly, 

a  =  JsV^t (2) 

Observe  carefully  that  the  tangents  to  the  hodograph  and  to 
the  path  at  corresponding  points  P  and  P'  are  not  parallel  or  in 

*  As  explained  in  the  appendix  referred  to,  the  fixed  point  A'  from 
which  s'  is  measured  is  to  be  taken  so  that  P'  moves  in  the  positive  5' 
direction  and  the  arrow  on  the  tangent  (giving  the  sense  of  the  vector) 
points  in  the  same  direction. 


§1] 


VELOCITY  AND  ACCELERATION. 


187 


any  way  related  as  to  direction,  but  the  vector  representing  the 
acceleration  always  points  from  P  toward  the  side  of  the  tan- 
gent on  which  the  path  lies. 

EXAMPLE. 

I.  What  is  the  acceleration  of  a  point  describing  a  circle  of 
radius  r  with  constant  speed  vt 

Solution:  Let  AB  (fig.  160)  be  the  circular  path  and  P  the 
position  of  the  moving  point  at  any  instant.     When  P  is  dX  A 


Fig.  160. 

the  velocity  is  represented  by  the  vector  0'A\  when  at  P  by  0'P\ 
when  at  B  by  O'B' ,  etc.  Hence  the  hodograph  is  the  circular 
arc  A'P'B',  its  radius  being  v.  Now  the  angles  PGA  and  P'O'A' 
are  equal,  and  since  the  former  equals  s/r  and  the  latter  s' /Vy 


Hence 


s/r  =  s'/Vy     or     s'^sv/r. 
ds'/dt  =  {ds/dt)v/r=vyr. 
a  =  v^/r. 


Since  ds' ldt  =  a^ 


The  direction  of  the  acceleration  of  P  is  that  of  the  tangent 
to  the  hodograph  at  P' .  Since  this  tangent  is  parallel  to  the 
normal  to  the  path  at  P, 

the  acceleration  of  a  point  describing  a  circle  with 
constant  speed  is  directed  along  the  radius  of  the 
circle  drawn  to  the  moving  point,  and  its  value 
equals  the  speed  squared  divided  by  the  length  of 
the  radius. 


1 88  "         CURyiLINEAR  MOTION.  [Chap.  VIII. 


§  II.     Resolutions  of  Velocities  and  Accelerations. 

189.  Components  and  Resultant  of  Velocities  or  Accelerations 
Defined. — Velocities  and  accelerations  are,  as  defined  in  arts. 
186  and  188,  vector  quantities;  they  may  therefore  be  resolved 
and  compounded.  The  velocity  (or  acceleration)  represented 
by  the  sum  of  the  vectors  representing  any  number  of  velocities 
(or  accelerations)  is  called  the  resultant  of  those  velocities  (or 
accelerations).  The  velocities  (or  accelerations)  represented  by 
the  components  of  a  vector  representing  any  velocity  (or  accel- 
eration) are  called  the  components  of  that  velocity  (or  accelera- 
tion) .     * 

190.  Axial  Components  of  a  Velocity. — These  components  are 
parallel  to  three  rectangular  axes  (it,  y,  and  2),  and  will  be  de- 
noted by  Vx,  Vy,  and  Vg  respectively. 

Let  V  denote  a  velocity,  its  direction  angles  with  the  x,  y, 
and  z  axes  being  a,  /5,  and  y  respectively.  Then  the  x,  y, 
and  z  components  equal  respectively  v  cos  a,  v  cos  /?,  and  v  cos  y. 
Since  v  =  ds/dt,  cos  a=dx/ds,  cos  l^  =  dy/ds,  and  cos  y  =  dz/ds, 

Vx  =  dx/dt,    Vj,  =  dy/dt,    v.  =  dz/dt.      .     .     .     (i) 

These  equations  show  that  the  x,  y,  and  z  velocity  components 
of  a  moving  point  are  respectively  the  time-rates  at  which  its 
distances  from  the  yz,  zx,  and  xy  planes  change,  and  that  they 
-equal  respectively  the  velocities  of  the  projections  of  the  moving 
point  on  the  axes  x,  y,  and  z. 

191.  Tangential  and  Normal  Components  of  Velocity. — These 
•components  are  parallel  to  the  tangent  and  normal  to  the  path 
at  the  point  where  the  velocity  is  resolved.  They  will  be  de- 
noted by  Vf.  and  Vn  respectively.  Since  the  velocity  at  any  point 
in  the  path  is  directed  along  the  tangent  to  the  path  at  that 
point, 

Yt  =  v     and     v„  =  o.       .....     (i) 

192.  Axial  Components  of  Acceleration. — These  components 
are  parallel  to  three  coordinate  axes  (x,  y,  and  z) ,  and  will  be  de- 
noted by  a^,  ay,  and  a^  respectively.  For  simplicity,  the  deduc- 
tion of  the  expressions  for  the  components  is  limited  to  the  case 


§  II.]      RESOLUTIONS  OF  VELOCITIES  AND  ACCELERATIONS,        189 

of  plane  paths  with  the  coordinate  axes  in  the  plane  of  the  path, 
but  the  discussion  can  be  extended  to  the  general  case  (resolu- 
tion of  the  acceleration  of  a  point  along  three  axes,  the  path 
being  tortuous). 

Let  PQ  (fig.  161)  be  the  path  of  a  moving  point  and  P'Q'  the 
hodograph  of  the  motion,  P  and  P'  being  any  two  corresponding 


(a)  (6) 

Fig.  161. 

points  on  the  path  and  hodograph  respectively.  Also  let  v  and 
a  denote  the  magnitude  of  the  velocity  at  P  and  its  angle  with 
the  X  axis  respectively.  Then  the  polar  coordinates  of  P'  are 
V  and  a,  and  its  rectangular  coordinates  are  v  cos  a  and  v  sin  a, 
or  Vx  and  Vy',  hence  the  coordinate  axes  in  fig.  161(6)  are  marked 
O'V^  and  O'Vy  instead  of  OX  and  OY. 

As  shown  in  art.  188,  the  acceleration  at  P  is  directed  along 
the  tangent  to  the  hodograph  at  P'\  hence  P'a  may  represent 
the  acceleration.  Now  if  a,  a^,  and  ay  denote  the  acceleration 
and  its  x  and  y  components  respectively,  it  follows  from  the 
figure  that 


ar,  =  a  cos  a' 


and 


a  sm  a' 


and  from  the  analogy  between  the  two  parts  of  the  figure, 

cos  a^  =  dvx/ds^     and     sin  a' =  dvy/ds\ 

ds'  being  the  length  of  the  infinitesimal  arc  on  the  hodograph  at 
P'.     Also,  from  art.  188,  a  =  ds' /dt\   hence 

a^  =  {ds'/dt){dv^/ds')  =dvjdt, 

and  ay  =  (ds'/dt)  {dvy/ds')  =  dvy/dt. 

In  the  general  case  (resolution  into  three  components)  it  can  be 


ipo  CURyiLlhlEAR  MOTION.  [Chap.  VIII.  | 

shown  that  the  z  component  of  the  acceleration  (a^)  equals  i 
dvz/dt\  hence  ^  j 

ax  =  dVx/dt  =  dVdt^  )  \ 

aj,  =  dvj,/dt  =  dVdt';  V (i)  i 

'       a,  =  dv,/dt  =  dVdtl  )    .  \ 

The  equations  state  that  the  x,  y,  and  z  components  of  the«; 
acceleration  of  a  moving  point  P  equal  the  time-rates  of  the  | 
X,  y,  and  z  components  of  its  velocity;  also  that  they  equal  j 
the  accelerations  of  the  projections  of  the  point  on  the  x,  y^  \ 
and  z  axes  respectively.  ^  j 

193.  Tangential  and  Normal  Components  of  Acceleration. —  \ 
These  are  components  whose  directions  are  parallel  to  the  '\ 
tangent  and  normal  to  the  path  at  the  point  where  the  accel-  i 
eration  is  resolved;  they  will  be  denoted  by  a^  and  a„  re-  \ 
spectively.     For  simplicity,    the    deduction    below   is   limited   ■ 

to  the  case  of  a  plane  path,  but  it  i 
might  be  extended  to  the  general  i 
case.  Let  a  denote  the  acceleration  ; 
of  a  moving  point  when  it  arrives  at  1 
P  (fig.  162).  Since  the  values  of  the  i 
tangential  and  normal  components  ] 
^^^'  ^^^-  of  the  acceleration  are  independent   ' 

of  the  axes  of  reference,*  these  may  be  chosen  parallel  to  the  \ 
tangent  and  normal  at  P.  This  is  done  solely  for  simplicity  in  ; 
the  deduction  of  the  values  of  at  and  a^.  With  axes  thus  chosen,  ■] 
it  is  plain  that  the  tangential  and  x  components  and  the  normal  • 
and  y  components  are  equal,  or  \ 

at  =  aj,  =  dv:,/dt,     an=ay=d^y/dt^.  ] 

Also  d^y/dt^  =  {d^y/dx^){dxydt^)  ={d^y/dx^)v^^;  and  d^y/dx^  \ 
equals  the  curvature  of  the  path  at  P.  Denoting  the  radius  of  j 
curvature  at  P  by  p,  and  since  Vx  =  Vt  =  v,  1 


a,  =  dv/dt  =  dVdt',     and     a„=vV^.     .     .     .     (i) 

*  This  independence  may  be  explained  thus:  From  fig.  161  it  is  plain 
that  at  =  a  cos  (a  — a')  and  an=a  sin  (a  — a').  Now  both  a  and  (a  — a') 
are  independent  of  the  coordinate  axes  and  hence  at  and  an  are  also. 


§  II.]      RESOLUTIONS  OF  VELOCITIES  ^ND  ACCELERATIONS.       191 

These  equations  state  that  the  tangential  and  normal  accelera- 
tions at  P  (any  point  in  the  path)  equal  respectively  the  time- 
rate  (of  change)  of  the  speed  at  P  and  the  square  of  the  speed 
divided  by  the  radius  of  curvature  at  P, 

If  the  moving  point  travels  in  a  circle  of  radius  r  and  with 
a  constant  speed  v,  then  dv/dt  =  o  and 


vVr. 


(2) 


EXAMPLES. 

I.  A  point  P  (fig.  163a)  starts  from  a  point  X  and  moves  in 
a  circle  of  20  ft.  radius  and  so  that  the  distance  described  (in 
feet)  equals  twice  the  cube  of  the  time  (in  seconds)  after  start- 


er) 


Fig.  163. 


ing.     Compute  the  value  of  v,  v^,  v^,  v^,  Vy,  a,  a^,  a„,  a^^,  and  a^, 
I-   when  t  =  2  sees. 
I  Solution:  From  the  equation  of  motion,  5  =  2/'; 

^  .'.ds/dt  =  6t^  =  v,     dh/dt^=i2t  =  at;     also,     an  =  s6t*/20, 

i     Hence,  when  t  =  2  sees., 

1;  =  24  ft. /sec,  2;/  =  24  ft. /sec,     Vn  =  o, 

I  a^  =  24  ft./sec.2,     and     a„  =  28.8  ft./sec^. 

Since  ^=20  cos  d  and  y=2o  sin  d, 

dx/dt  =  -  20  sin  (9  dd/dt  =  v^,,     dy/dt  =  20  cos  d  dd/dt=-Vy, 
dH/dt''^  -20  sin  d  d''d/dt''-2o{dd/dtY  cos  ^  =  a^, 
and 

d'^y/dt''  =  20  cos  6  d^'O/df  -  2o{dd/dty  sin  6  =  ay. 

Since  d  =  s/2o  =  o.it^,  d6/dt  =  o.;^t^,  and  d^d/dt^  =  o.6t.  By  means 
of  these  values  and  the  expressions  for  Vx,Vy,  a^,  and  Gy,  we  find 
that  when /  =  2  sees.. 


192  CURVILINEAR  MOTION,  [Chap.  VIIL 

Vx=  —11.21     and     Vy=    16.7  ft. /sec, 
<^x=— 37-2       and     a2,= -*-.3.9  ft./sec.^. 

The  determination  of  the  magnitude  and  direction  of  a  is  left  to 
the  student;  he  should  also  plot  to  scale,  in  the  figure,  all  the 
computed  values. 

2.  Discuss  the  motion  of  the  centre  of  gravity  of  a  projectile 
neglecting  the  resistance  of  the  air  if  the  velocity  and  angle  of 
projection  are  v'  and  e  respectively  (see  fig.  1636). 

Solution:  It  is  shown  in  art.  240  that  the  acceleration  of  the 
centre  of  gravity  at  each  instant  is  just  like  that  of  a  freely  fall- 
ing body,  i.e.,  equal  to  g  and  vertically  downward;  hence 

a^  =  o     and     ay=-g (i) 

Since  d'^x/dt'^  =  o,Vx  =  C^  and  x  =  C^t-]rC2,  C^  and  C^  being  con- 
stants  of  integration.  Since  i'^  is  constant  during  the  entire 
motion,  it  is  equal  to  its  initial  value,  i.e.,  C^  =  v'  cos  s,  and  since 
x  =  o  when  /  =  o,  C2  =  o ;  hence 

Vx  =  v'  cos  £     and     :\[:  =  z/'cos  e-t (2) 

Since  d^y/dt^  =  —g,  "^j/  =  —  ^^  +  Q  and  y=  —  igt^  +  C^t  +  C4, 
C3  and  C4  being  constants  of  integration  determinable  like  C^ 
and  C2  from  *  *  initial  conditions . ' '  These  are  y  =  o  and  Vy  =  'i/  sin  e 
when  /  =  o ;  hencp  Q  =  v'  sin  e  and  C^  =  o,  and  therefore 

Vy=  —gt  +  v'  sin  e     and     y=  —^gt'^-{-v'  s\n  e-t,     .     (3) 

3.  Show  that  the  path  is  parabolic  and  that  its  equation  is 

y  =  x  tan  £—gx'^/2{i/  cos  £f. 

4.  Deduce  expressions  for  the  range  and  greatest  height 
reached,  and  show  what  values  of  e  make  them  a  maximum. 

§  III.     Relativity  of  Motion. 

194.  Path,  Displacement,  Velocity,  and  Acceleration  of  a 
Point  Relative  to  a  Body. — Position  of  a  point  can  be  specified 
only  by  means  of  a  set  of  reference  axes  or  some  other  equiva- 
lent base.  When  we  speak  of  the  position  of  a  point  with 
respect  to  a  body  we  would  specify  that  position  relative  to 
some  reference  base  in  the  body;  thus  to  specify  the  position 
of  a  bug  on  a  drawing-board,  say,  we  might  describe  it  as  being 


§111.]  RELATIVITY  OF  MOTION.  193 

on  the  upper  surface  of  the  board,  two  feet  above  the  lower  edge 
and  three  feet  from  the  left  edge.  The  fact  that  specification  of 
position  is  necessarily  relative  makes  the  path,  displacement, 
velocity,  and  acceleration  of  a  moving  point  also  relative,  as 
can  be  shown  clearly  by  illustration. 

Imagine  a  drawing-board  with  a  sheet  of  paper  lying  upon  it, 
and  a  bug  upon  the  paper.  Suppose  that  the  paper  is  slid  over  the 
board  and  that  the  bug  walks  about  on  the  paper  and  punches 
holes  rapidly  through  the  paper  into  the  board.  The  succes- 
sions of  holes  in  the  paper  and  board  mark  out  his  paths  rela- 
tive to  the  paper  and  board  respectively.  Evidently  these 
paths  would,  in  general,  be  very  dissimilar. 

Let  Pj  and  B^  denote  the  holes  punched  into  the  paper 
and  board  at  any  instant,  and  P2  and  B^  those  punched  at  a 
later  instant.  Then  the  vectors  P^B^  and  P^B^  are  the  dis- 
placements of  the  bug  relative  to  the  paper  and  the  board 
respectively  for  the  interval  of  time.  Evidently  these  two 
displacements  would,  in  general,  be-  very  dissimilar.  The 
rates  at  which  the  displacements  relative  to  the  paper  and 
the  board  occur  are  the  velocities  of  the  bug  relative  to  the 
paper  and  the  board  respectively.  Since  these  displacements  for 
any  interval  of  time  are,  in  general,  unlike,  the  velocities  at  any 
instant  are  also  unlike.  Furthermore,  the  changes  or  incre- 
ments in  the  velocities  for  any  interval  are  unlike  also. 

The  rates  at  which  the  velocities  relative  to  the  paper  and 
the  board  change  are  the  accelerations  of  the  bug  relative 
to  the  paper  and  the'  board  respectively.  Since  the  changes 
in  the  velocities  for  any  interval  of  time  are,  in  general,  unlike; 
the  accelerations  at  any  instant  are  also  unlike. 

195.  Path,  Displacement,  Velocity,  and  Acceleration  of  a 
Point  Relative  to  Another  Point. — By  path,  displacement,  ve- 
locity, and  acceleration  of  a  point  relative  to  a  second  point  is 
meant  the  path,  displacement,  velocity,  and  acceleration  of  the 
point  relative  to  a  set  of  axes  having  fixed  directions  through  the 
second  point. 

If  a  bug  walks  about  on  a  sheet  of  paper  whose  edges  are 
kept  fixed  in  direction,  then  its  motions  relative  to  the  paper 
and  any  point  of  the  same  are  alike;  but  if  the  edges  are  not 
kept  fixed  in  direction,  then  those  motions  are  unlike.  Thus 
imagine  two  sheets  of  paper  pivoted  together  at  some  point  O, 


194 


CURyiLINEAR  MOTION. 


[Chap.  VIII. 


that  they  are  slid  over  a  flat  surface  in  any  way  except  that 
the  edges  of  the  lower  sheet  remain  fixed  in  direction,  and  that 
a  punching  bug  walks  about  on  the  upper  sheet.  The  holes 
punched  into  the  lower  sheet  mark  the  path  of  the  bug  relative 
to  that  sheet  and  to  any  point  of  the  sheet.  The  holes  punched 
in  the  upper  sheet  mark  the  path  relative  to  that  sheet;  this 
path  would  in  general  be  unlike  the  first,  which  is  the  path  of 
the  bug  relative  to  the  point  O  of  the  upper  sheet. 

196.  Relation  between  the  Velocities  and  the  Accelerations 
of  (a)  Two  Points  and  (b)  Three  Points. — (a)  Proposition. — 
The  velocities  and  the  accelerations  of  two  points  relative  to 
each  other  are  equal  and  opposite. 

Proof   for   uniplanar   motion   of  the   points:     Imagine   two 
sheets   of  paper  being  shifted  about   on  a  flat  surface,  their 
edges  remaining  fixed  in  direction.    Also  imagine  punching  bugs 
fixed  at  the  middles  of  the  sheets  and  call  the  one  on  the  upper 
sheet  A  and  the  other  B ;  then  the  suc- 
cessions of  holes    punched  by  A   and  B 
(into    the    lower    and    upper    sheets    re- 
spectively) mark  the  paths  of  A  relative 
to  B  and  B  relative  to  A.     Let  (a)  and  (6) 
(fig.   164)  represent   the   positions  of  the 
sheets  at   the  beginning  and  end  of  any 
motion;    then   A^  and  B\    are    the    first 
positions  of   the  bugs  A   and  B,  and  A.^ 
and    B2    are    their    last    positions.       If 
A^A^B^B'    and    B^B^A^A'    are    parallelo- 
grams, then  B'  and  .4'  are  the  final  posi- 
tions of  the  first  holes  punched  into  the 
Fig.  164.  upper     and     lower     sheets     respectively, 

and  hence  the  displacements  of  A  relative  to  B  and  B  relative 
to  A  are  the  vectors  A'A^  and  B'B.^.  Since  the  sides  A.^B' 
and  B^A'  of  the  quadrilateral  ^g^'-^z-^'  are  equal  and  parallel, 
the  mentioned  vectors  (and  displacements)  are  equal;  plainly 
they  are  opposite. 

It  follows  that  the  displacements  per  unit  time,  and  hence 
the  velocities,  of  the  two  points  are  equal  and  opposite.     The 


§111. 


RELATI^^ITY  OF  MOTION. 


195 


velocities  being  at  each  instant  equal  and  opposite,  it  follows 
that  the  changes  in  those  velocities  for  any  interval  and  the 
velocity  changes  per  unit  time  (hence  the  accelerations  also) 
must  be  equal  and  opposite. 

(6)  Proposition. — The  velocity  (and  acceleration)  of  A  rela- 
tive  to  B  plus  that  of  B  relative  to  C  equals  that  of  A  relative 
to  C. 

Proof  for  uniplanar  motion  of  the  three  points:  Imagine  a 
sheet  of  paper  slid  about  on  a  drawing-board  which  is  also 
slid  about  on  a  fiat  surface,  the  edges  of  the  paper  and  board 
being  fixed  in  direction.  Imagine  also 
a  punching  bug  walking  about  on  the 
paper;  let  it  be  A,  a  corner  of  the 
paper  B,  and  a  corner  of  the  board  C. 
Suppose  that  at  the  beginning  of  some 
interval  the  board,  paper,  and  bug  are  at 
CjCj,  B^h^y  and  A^  (fig.  165)  respectively, 
and  at  the  end  of  that  interval  they 
are  at  C^c^,  BJD^y  ^^^  ^2-  I^  C^C^BJB^ 
B'B,A^A\  and  B^B'A'A''  are  parallelo- 
grams, it  follows  respectively  that  B'B^ 
is  the  displacement  of  B  relative  to  C; 
that  A'  is  the  final  position  of  the  first 
hole  punched  in  the  board,  and  hence  the  displacement  of  A 
relative  to  C  is  A' A 2',  and  that  ^"  is  the  final  position  of  the 
first  hole  punched  in  the  paper,  and  hence  A^'A2  is  the  dis- 
placement of  A  relative  to  B.  It  is  plain  from  the  figure  that 
vectoHally  A'^A^  +  B'B^^A^A^,  i.e.,  the  displacement  of  A 
relative  to  B  plus  that  of  B  relative  to  C  equals  that  of  A  rela- 
tive to  C. 

It  follows  that  the  displacement  per  unit  time,  and  hence  the 
velocities,  are  similarly  related.  The  velocities  at  each  instant 
being  thus  related,  it  follows  that  the  changes  in  those  velocities 
in  any  interval  and  the  velocity  changes  per  unit  time  (hence 
the  accelerations  also)  are  so  related. 

197.  Meaning  of  Composition  of  Motions. — According  to  the 
preceding  article  the  displacement  of  the  bug  (of  the  first  illustra- 


FiG.   165. 


196 


CURVILINEAR  MOTION, 


[Chap.  VIII. 


tion  of  art.  194)  with  respect  to  a  point  C  on  the  board  may  be 
found  by  compounding  its  displacen;ent  with  respect  to  a  point 
B  on  the  paper  and  that  of  B  with  respect  to  C.  Hence  we  say 
that  the  motion  of  a  point  A  with  respect  to  a  point  C  may  be 
regarded  as  consisting  of  the  motion  of  A  with  respect  to  B  and 
that  of  B  with  respect  to  C. 

EXAMPLES. 

1.  A  man  rows  a  boat  in  a  stream  whose  velocity  is  3  miles 
per  hour,  so  that  his  velocity  relative  to  a  floating  chip  is  5  mi.- 
per-hr.  "straight  across"  the  stream.  Determine  his  absolute 
velocity.* 

2.  Two  locomotives  run  at  speeds  of  40  and  50  mi.-per-hr.  on 
two  different  tracks,  that  of  the  first  locomotive  being  east  and 
west  and  the  other  northeast  and  southwest.  If  both  locomotives 
run  eastward,  what  is  the  velocity  of  each  relative  to  the  other? 


§  IV.     Composition  of  Simple   Harmonic   Motions. 

198.  Mechanism  for  Compounding  Simple  Harmonic  Motions> 

— Fig.  166(a)  represents  two  cranks  with  their  slotted  sliders, 
the  crank-shaft  of  one  turning  in  a  bearing  fixed  upon  the  slider 
of  the  second.     If  both  cranks  be  turned  uniformly,  the  lower 


Fig.  166. 

slider  S' executes  a  s.h.m.  relative  to  the  support  of  the  mechan- 
ism, the  upper  shder  S"  executes  a  s.h.m.  relative  to  the  lower 

*  Motion  referred  to  points  on  the  earth  is  often  called  for  convenience 
"  absolute  motion,"  and  the  corresponding  velocities  and  acceleration  are 
also,  called  absolute. 


§  IV.]        COMPOSITION  OF  SIMPLE  HARMONIC  MOTIONS. 


197 


slider,  and  its  mbtion  relative  to  the  fixed  support  is  compounded 
of  these  two  harmonic  motions.  In  theory  at  least,  a  third  crank 
and  slider  could  be  mounted  on  the  second  slider,  a  fourth  crank 
and  slider  on  the  third  slider,  etc.  Then  if  all  cranks  were  turned 
uniformly,  the  motion  of  the  last  slider  would  be  compounded 
of  the  simple  harmonic  motions  executed  by  the  several  sliders 
relative  to  the  supports  of  their  cranks. 

When  the  sliders  are  in  line  the  s.h.m.'s  are  described  as 
* 'along  the  same  line"  or  "collinear,"  and  when  the  sliders  are 
inclined  to  each  other  the  s.h.m.'s  may  be  described  as  "oblique  " 
or  "  non-parallel." 

199.  Composition  of  Two  CoUinear  Simple  Harmonic  Motions 
of  the  Same  Period.* — Let  fig.  166  represent  the  position  of  the 
mechanism  at  any  time  t  of  the  motion.  We  will  consider  the 
motion  of  the  projection  of  C2  on  OF,  i.e.,  V. 

From  0  draw  vectors  OP'  and  OP"  to  represent  the  cranks 
as  shown,  and  complete  the  parallelogram  OP'P"P\  then  OP  is 
the  sum  of  the  vectors  representing  the  cranks.  It  is  plain  from 
the  figure  that  V  is  also  the  projection  of  P  on  the  line  OY. 
Since  the  periods  of  the  two  s.h.m.'s  are  the  same,  the  angle 
P'OP"  remains  constant  during  the  motion;  hence  OP  is  con- 
stant in  length  and  turns  uniformly,  i.e.,  describes  a  circle  at  a 
uniform  rate.     Therefore  the  motion  of  V  is  simply  harmonic. 

The  amplitude  of  the  motion  of  Y  is  OP,  the  phase  is  XOP, 
the  period  is  the  same  as  that  of  the  given  s.h.m.'s.  The  epoch 
can  be  found  from  the  figure  as  follows :  Turn  the  parallelogram 
back  to  its  position  when  ^  =  o ;  let  Pq  be  the  corresponding  posi- 
tion of  P,  then  XOPq  is  the  desired  epoch. 

Values  of  the  amplitude  and  epoch  can  be  computed  from  the 
figure ;  thus  let 


a^  denote  the  amplitude    of  the  first      s.h.m.. 


a^       " 

it         <<         << 

"     second     " 

£1 

"     epoch              " 

"    first 

^2 

<<                    H                                             <( 

"    second    ** 

y^     " 

"  displacement  " 

"    first 

y2     " 

li             <<              << 

"    second     " 

*  Only  motions  of  the  same  period  are  compounded  herein. 


»98  CURI^ILINEAR  MOTION,  [Chap.  Vlll. 

and  2 ;r/ 6;  their  common  period; 

then  y^  =  a^sm{ojt+z^     and     %  =  a^  sin  {cot  ■\- ^2) . 

Since   the   motion  of  V  is   simply  harmonic,  period   equal   to 
27z/oj,  its  displacement  is  given  by 

y  =  a  svn  {wt -\- e) , 

1  and  e  being  the  amplitude  and  epoch  respectively.     It  can  be 
shown  from  the  figure  that 

and         tan  e  =  (a^  sin  s^  +  a^  sin  e2)/(«i  cos  s^  +  ag  cos  £2) • 

200.  Resolution  of  a  Simple  Harmonic  Motion  into  Two  Com- 
ponents CoUinear  with  it. — Let  OP  (fig.  166)  represent  the  crank 
of  any  s.h.m.  in  the  line  OY ,  and  let  OP'  and  OP"  be  the  cranks 
of  two  others  in  the  same  line.  According  to  the  preceding  ar- 
ticle the  resultant  of  the  second  two  gives  the  first  and  the  latter 
are  therefore  called  components  of  the  first. 

Obviously  a  s.h.m.  may  be  resolved  into  many  pairs  of  com- 
ponents, for  many  parallelograms,  as  OP'P"P,  can  be  drawn  on 
the  same  diagonal  OP,  and  the  sides  OP'  and  OP"  of  each  rep- 
resent the  cranks  of  a  pair  of  components. 

Special  Case. — Resolution  into  two  components  which  differ 

90°  in  phase,  the  epoch  of  one  being  zero.     Let  OP  (fig.  167) 

represent  the  crank  of  the  s.h.m.  to  be 

resolved  when  /  =  o;    then  XOP  is  the 

epoch  of  that  motion  and  OP'  and  OP" 

X    are  the  cranks  of  the  component  motions 

when  ^  =  o ;  the  epoch  of  one  component 

Fig.  167.  (crank  OP')  is  zero  and  that  of  the  other 

is  90°.     Also  if  OP  =  a  and  XOP=--e,  the  equation  of  the  given 

s.h.m.  is 

y  =  a  sin  (ojt+e), 

and  those  of  the  first  and  second  components  are 

y'  =  a  cos  £-sin  cot, 

and  y"  =  a  sin  £(sin  cot  4-90°)  =  a  sin  e  •  cos  (ot. 


IV.]        COMPOSITION  OF  SIMPLE  HARMOMIC  MOTIONS. 


199 


201.  Composition  of  Many  Collinear  Simple  Harmonic  Motions 
of  Equal  Periods. — It  follows  from  art.  199  that  the  resultant  of 
three  or  raore  collinear  s.h.m.'s  of  equal  period  is  simply  har- 
monic and  of  that  period.  For  the  first  two  can  be  replaced  by  a 
single  s.h.m.,  and  in.  turn  this  one  and  the  third  can  be  replaced 
by  a  single  s.h.m.,  etc. 

To  obtain  the  "crank"  of  the  resultant  motion,  add  the  vec- 
tors representing  the  several  cranks  in  their  positions  at  any 
instant ;  the  sum  represents  the  crank  of  the  resultant  motion 
in  its  corresponding  position.  Thus,  let  0P\  P'P",  P"P'",  etc. 
(fig.  168),  represent  the  cranks  in  their  simultaneous  positions; 


Fig.  168. 

then  OP  represents  the  length  and  direction  of  the  crank  of  the 
resultant  motion  at  the  same  instant. 

The  epoch  can  be  determined  by  turning  the  polygon  about 
0  until  P'  falls  into  its  position  when  /  was  zero.  Let  Pq  be  the 
corresponding  position  of  P;  then  XOP^  is  the  epoch  sought. 

EXAMPLES. 

1.  Compound  two  collinear  s.h.m.'s  whose  amplitudes  and 
periods  are  equal,  their  phases  differing  (a)  by  90°,  (6)  by  180*^, 
(c)  by  270°. 

2.  Resolve  the  s.h.m.  whose  equation  is  y  =  4  sin  {27rt  +  ^o°) 
into  two  collinear  with  it,  their  phase  difference  being  45°  and 
the  epoch  of  one  being  zero. 

3.  Compound  three  collinear  s.h.m.'s  whose  amplitudes  and 
periods  are  equal,  their  phases  differing  by  120°. 


200 


CURVILINEAR  MOTION. 


[Chap.  VIII. 


202.  Composition  of  Two  Simple  Harmonic  Motions  in  Lines 
at  Right  Angles. — Imagine  the  two  sliders  of  fig.  i66  turned  at 

right  angles  as  represented  in  fig.  169; 
then  if  the  absolute  motion  of  the 
lower  one  is  simply  harmonic  and  the 
motion  of  the  upper  one  relative  to 
the  lower  is  also  simply  harmonic,  the 
absolute  motion  of  the  upper  is  com- 
pounded of  two  s.h.m.'s  at  right 
angles. 

For  simplicity,  we  choose  the  ori- 
gin of  time  so  that  the  epoch  of  the 
first  s.h.m.  is  zero  and  consider  the 
motion  of  a  point  P  at  the  middle  of 
the  slot  of  the  second  slider.  With 
Fig.  169.  notation  and  axes  as  in  the  figure,  the 

X  and  y  coordinates  of  P  are 

x  =  a^s>mo)t     and     ;y  =  a2  sin  (0;/  +  ^).    .     .     .     (i) 

The  equation  of  the  path  of  P  is  found  from  these  by  eliminating 
/;  thus  we  find  that 


2  cos 


y 
^:^+— 2=sm 

02 


(2) 

the   motion  has  been 


This  represents    an    ellipse    and    hence 
called  **  elliptic  harmonic  motion." 

The  path  of  a  point  whose  motion  is  the  resultant  of  two 
s.h.m.'s  in  lines  at  right  angles  (or  inclined)  can  be  traced 
graphically  by  plotting  on  the  two  lines  simultaneous  values  of 
the  displacements  due  to  the  component  motions  and  then  fix- 
ing the  position  of  the  point  from  the  displacements.  Thus  let 
the  circles  of  fig.  170  be  the  circles  of  reference  of  the  two  com- 
ponent s.h.m.'s,  and  let  VOa^  be  the  epoch  of  the  horizontal 
and  XOh^  that  of  the  vertical  component  motion ;  then  the  dis- 
placements in  the  two  motions  when  ^  =  0  are  respectively  Ox^ 
and  Oy^,  and  the  position  of  the  point  describing  the  elliptic 
motion  is  Cq.  The  constructions  for  the  other  points  on  the 
ellipse  should  be  obvious  from  the  figure. 

Special  Cases. — (i)  If  the  phases  of  the  component  motions 


§  IV.]        COMPOSITION  OF  SIMPLE  HARMONIC  MOTIONS. 


20T 


are  the  same  or  differ  by  i8o°,  sin  ^  =  o;    hence   according  to 

eq.  (2)  the  path  is  a  straight  line.     The  resultant  motion  is  a 

I   s.h.m.,  its  period  equals  that  of  the  given  motions,  its  phase  is 

r  ♦ 

Y 

4 


Fig.  170. 
the  same  as  that  of  one  or  both  of  the  given  motions,  and  its 
amplitude  equals  (Gti*  +  a2^)*-     Prove. 

(2)  If  the  amplitudes  of  the  component  motions  are  equal 
and  they  differ  in  phase  by  90°,  a^  =  a2,  cos^  =  o,  and  sin  ^  =  1; 
hence  according  to  eq.  (2)  the  path  is  a  circle  whose  radius 
equals  a^^a^.  The  circular  motion  is  "uniform,"  of  a  period 
equal  to  that  of  the  component  motions.     Prove. 

203.  Resolution  of  a  S.H.M.  in.to  Two  Components  at  Right 
Angles  to  Each  Other.— Let  OZ  (fig.  171)  be  the  path  of  the 
motion  to  be  resolved,  O  being  the  centre,  and 
let  the  displacement,  amplitude,  period,  and 
epoch  be  0,  a,  2tzJ uj,  and  e  respectively;  then 

2  =  a  sin  (a;/+e). 

The  %■  and  y  components  of  this  displacement 
are  (see  the  figure), 

%  =  {a  cos  a)  sin  (wZ  +  e) 

and  :V  =  («sina:)  sin  (a>/H-e);  Fig  171 

hence  the  periods  and  epochs  of  the  components  are  the  same 


202 


CURVILINE/IR   MOTION. 


[Chap.  VIII- 


as  of  the  given  motion,  and  their  ampHtudes  are  respectively 
a  cos  a  and  a  sin  a.  , 

204.  Composition  of  More  than  Two  S.H.M.*s  Not  Collinear. — 

According  to  the  preceding  article  each  s.h.m.  may  be  resolved  and 
replaced  by  two  components  along  two  axes  x  and  y.  Accord- 
ing to  art.  201  all  the  components  in  the  x  axis  can  be  com- 
pounded into  a  single  s.h.m. ,  and  those  in  the  y  axis  also.  Accord- 
ing to  art.  202  these  two  s.h.m. 's  compound,  in  general,  into  an 
elliptic  harmonic  motion. 

Special  Case. — Three  s.h.m. 's  in  lines  inclined  120°  to  each 
other  of  equal  amplitudes  and  periods  but  differing  120°  in 
phase,  compound  into  a  uniform  circular  motion. 

Proof :  Let  z'z"  and  z'"  denote  simultaneous  displacements  in 
the  component  motions  and  a  and  27r/a;  their  common  amplitude 

and  period;  then 

z'  =  a  sin  cot, 
z"  =  a  sin  (o;/-f- 120°), 
z'"  =  a  sin  (o;/  +  240°). 
Let  OA,  05,  and  OC  (fig.  172)  be 
the  paths  of  the  component  mo- 
tions, 0  being  the  centre  of  each, 
%' ,  x" ,  and  x'"  the  x  components 
of  the  displacements,  and  y' ,  y" y 
and  y'"  their  y  components.    Then,, 
according  to  the  preceding  article,. 


Fig.  172. 

x' ^a  sin  (to/), 
x"  ==a  cos  1 20° -sin  (<w/  +  i2o°). 


y  =0, 

y"  =  a  sin  120° -sin  {cut -\- 120°), 
'"  —  a  sin  240°  •  sin  {wt-\-  240°) ; 
hence  the  sums  of  the  x  and  y  components  (which  call  x  and  y) 
are 

x-^^asm  (x)t,  and  ;v  =  |a  cos  a;/ =  f  sin  (0^^  +  90°) . 
These  two  equations  show  that  the  given  motions  are  equivalent 
to  two  s.h.m.'s  in  the  coordinate  axes,  equal  as  to  amplitudes  and 
periods  but  differing  90°  in  phase.  According  to  special  case  (2 ) , 
art.  202,  these  two  motions  are  equivalent  to  a  uniform  motion 
in  a  circle  of  radius  |a,  its  period  being  equal  to  that  of  the 
given  motions. 


=  a  cos  240° -sin  (a>/ -1-240°),  y' 


CHAPTER  IX. 
MOTION  OF  A  RIGID  BODY. 

§  I.     Translation. 

205.  Translation  Defined. — A  translation  is  a  motion  in  which 
each  straight  line  of  the  moving  body  remains  fixed  in  direction. 
Thus  the  motions  of  the  coupling-rods  of  a  locomotive  which  runs 
on  a  straight  -  track  are  translations ;  also  the  motions  of  those 
on  a  locomotive  which  runs  on  a  transfer-table. 

206.  Motions  of  all  Points  of  a  Body  in  Translation  are  Alike. 
— Let  A  and  B  be  any  two  points  of  a  body  having  a  translatory 
motion,  A^  and  5' their  positions  at  a  given  instant,  and  A''  and 
5"  those  at  another  instant.  By  definition,  the  lines  A'B^  and 
A"B"  are  parallel,  and,  since  they  are  also  equal  in  length,  the 
figure  A'B'A"B"  is  a  parallelogram  and  A' A"  and  B'B"  are 
equal  and  parallel.  Hence  the  displacements  of  all  points  of 
the  moving  body  for  the  same  interval  of  time  (long  or  short) 
are  equal  in  magnitude  and  the  same  in  direction.  It  follows 
that  at  each  instant  the  velocities,  and  hence  the  accelerations, 
of  all  points  of  the  moving  body  are  exactly  alike. 

207.  Velocity  and  Acceleration  of  the  Body. — By  velocity 
and  acceleration  of  a  body  having  a  translatory  motion  is  meant 
the  velocity  and  the  acceleration  respectively  of  any  one  of  its 
points. 

§  II.     Rotation. 

208.  Rotation  Defined. — A  rotation  is  a  motion  in  which  one 
line  of  the  moving  body  or  of  its  extension  remains  fixed.  The 
fixed  line  is  called  the  axis  of  the  rotation. 

Obviously  all  points  of  the  moving  body  must  describe  cir- 
cles whose  centres  lie  on  the  axis  unless  the  axis  cuts  the  body; 
in  this  case  all  points  of  the  body  on  that  line  are  at  rest,  the 
others  describing  circles.     The  planes  of  the  circles  are  perpen- 

203 


S04  MOTION  OF  A  RIGID  BODY.  [Chap.  IX. 

dicular  to  the  axis,  and  any  plane  perpendicular  to  the  axis  may 
be  called  the  plane  of  rotation.  All  ^points  of  the  body  on  any 
line  parallel  to  the  axis  move  alike;  hence  the  motion  of  the  pro- 
jection of  the  line  on  the  plane  of  rotation  represents  that  of 
all  the  points,  and  the  motion  of  the  body  itself  is  represented  by 
the  motion  of  its  projection. 

209.  Angular  Displacement. — By  angular  displacement  of  a 
rotating  body  during  any  time  interval  is  meant  the  angle  de- 
scribed during  that  interval  by  any  line  of  the  body  perpendicular 
to  the  axis.  Obviously  all  such  lines  describe  equal  angles  in  the 
same  interval,  and  we  select  a  line  which  cuts  the  axis.  For  con- 
venience, angular  displacements  are  given  sign — positive  if  dur- 
ing the  interval  the  body  has  turned  counter-clockwise,  and 
negative  if  clockwise. 

Let  the  irregular  outline  (fig.  173)  represent  a  rotating  body, 
the  plane  of  rotation  being  that  of  the 
paper,  and  O  the  intersection  of  the 
axis  with  that  plane.  Let  P  be  any 
point  and  6  the  angle  XOP,  OX  being 
any  fixed  line  of  reference.  As  custom- 
arily, 6  is  regarded  as  positive  or  nega- 
tive according  as  OX  when  turned 
about  O  toward  OP  moves  counter- 
^^'  ^^^*  clockwise  or  clockwise.     If  d^  and  ^2 

denote  initial  and  final  values  of  6  corresponding  to  any  rota- 
tion, then  the 

angular  displacement  =  ^2  ~~  ^1  =  ^^' 

210.  Angular  Velocity. — The  angular  velocity  of  a  rotating 
body  is  the  time-rate  at  which  its  angular  displacement  occurs ; 
or,  otherwise  stated,  it  is  the  time-rate  at  which  any  line  of  the 
body  perpendicular  to  the  axis  describes  angle. 

The  time-rate  at  which  OP  (fig.  173)  describes  angle  or  the 
time-rate  (of  change)  of  ^  is,  as  shown  in  works  on  calculus  and 
in  Appendix  B,  dd/dt.     Hence  if  cj  denotes  angular  velocity, 

co  =  dd/dt (i) 

If  the  body  turns  uniformly,  6  is  a  uniform  variable,  and  its  time- 


§11.]  ROTATION.  2C5 

rate  is  Ad/ At,  Ad  denoting  the  angular  displacement  occurring  in 
the  interval  At.     Hence 

oj  =  Ad/At, (2) 

and  the  angular  velocity  is  constant. 

211.  Units  of  Angular  Velocity. — The  formulas  of  the  pre- 
ceding article  imply  as  unit  an  angular  velocity  corresponding 
to  a  unit  angular  displacement  in  each  unit  time,  the  velocity 
being  constant.  There  are  several  such  units ;  thus,  one  revolu- 
tion-per-second,  one  degree-per-hour,  one  radian-per-second,  etc. 
The  last  is  the  one  usually  used  herein.* 

212.  Sign  of  Angular  Velocity. — An  angular  velocity  must  be 
regarded  as  having  sign,  the  same  as  that  of  dO/dt  (and  of  Ad /At 
if  the  angular  velocity  is  constant).  Now  dO/dt  and  Ad /At  are 
positive  or  negative  according  as  d  increases  or  decreases  alge- 
braically ;   hence 

.  The  angular  velocity  of  a  rotating  body  at  any  in- 
stant is  positive  or  negative  according  a.s  it  is 
turning  in  the  counter-clockwise  or  clockwise 
direction  at  that  time. 

213.  Angular  Acceleration. — The  angular  acceleration  of  a 
rotating  body  is  the  time-rate  (of  change)  of  its  angular  velocity. 

If,  as  in  the  preceding,  o)  denotes  the  angular  velocity,  then 
the  general  expression  for  the  time-rate  of  the  angular  velocity 
is  dco/dt\  hence  if  a  denotes  the  angular  acceleration, 

a  =  &co/6X  =  &W/6X\    ......     (i) 

//  the  angular  velocity  changes  uniformly,  its  time-rate  is 
A  CO /At,  A  CO  denoting  the  increment  in  the  velocity  for  any  inter- 
val At;  hence 

a  =  Aoj/At (2) 

and  the  angular  acceleration  is  constant. 

2 14.  Units  of  Angular  Acceleration. — The  formulas  of  the  pre- 
ceding article  imply  as  unit  an  angular  acceleration  correspond- 
ing to  a  unit  angular  velocity  change  in  each  unit  time,  the  angu- 
lar  acceleration   being   constant.     One   revolution-per-second- 

*  For  dimensions  of  a  unit  angular  velocity,  see  Appendix  C. 


2o6  '  MOTION  OF  a   RIGID  BODY,  [Chap.  IX. 

per-second,  one  radian-per-second-per-second,  etc.,  are  such 
units.* 

215.  Sign  of  Angular  Acceleration. — An  angular  acceleration 
must  be  regarded  as  having  sign — the  same  as  that  of  dco/dt 
(and  of  A  CO /At  if  the  angular  velocity  changes  uniformly).  Now 
dw/dt  and  Aoj/At  are  positive  or  negative  according  as  oj  in- 
creases or  decreases  algebraically;   hence 

An  angular  acceleration  is  positive  or  negative  ac- 
cording as  the  angular  velocity  is  increasing  or  de- 
creasing (algebraically). 

216.  Velocity  and  Acceleration  of  Any  Point  of  a  Rotating 
Body. — Let  P  (fig.  173)  be  any  point  of  the  rotating  body  there 
represented,  let  r  denote  its  distance  from  the  axis  and  5  the 
length  of  the  arc  PqP.  Then  if  d  is  expressed  in  radians,  s=^rd; 
hence 

ds/dt  =  rdd/dt     and     dh/dt^  =  r  dW/di\ 

Now  ds/dt  is  the  velocity  of  P  (art.  186)  and  d^s/dt^  is  its  tan- 
gential acceleration  (art.  193);  hence,  if  as  heretofore  the  veloc- 
ity, the  acceleration,  and  its  tangential  and  normal  com- 
ponents be  denoted  hy  v,  a,  a^,  and  a„  respectively, 

y  =  Toj, (i) 

a^  =  ra,     a.n  =  Tco^y (2) 

and  a  =  rVa^  +  a;^;      . (3) 

OJ  and  a  denoting  respectively  the  angular  velocity  atid  acceler- 
ation of  the  body.  These  equations  show  that  the  velocity 
and  acceleration  of  a  point  in  a  rotating  body  are  proportional 
to  its  distance  from  the  axis. 

EXAMPLES. 

I.  Write  in  the  proper  place  belov/  the  signs  of  the  angular 
velocity  and  acceleration  of  a  body  which  rotates  as  follows : 

(a)  Clockwise, 

( 1 )  when  *  'getting  up  speed ,' '  sign  of  6;  is  .."'.,  oi  a  .  t . ; 

(2)  "      "slowing  down,"  "     "  "  "  .  .-.,    "  "  .^. . ; 

*  For  dimensions  of  an  angular  acceleration,  see  Appendix  C 


§  III.]  PLANE  MOTION,  .  207 

(h)  Counter-clockwise, 

(i )  when  "getting  up  speed,"  sign  oiujv&  .  ~. . ,   of  a  ... ; 
(2)       "      "slowing  down,"         "     """...,    "".... 

2.  Express  an  angular  velocity  of  n  rev.-per-sec.  in  rad.-per- 
sec.         ;^■^v 

3.  If  the  angular  velocity  of  a  wheel  changes  from  100  to  120 
rev.-per-min.  in  one-half  a  min.,  what  is  its  average  angular 
acceleration?  ^-^^  /r-^  c^c  "^ 

4.  A  wheel  is  set  rotating  in  such  a  manner  that  the  num- 
ber of  turns  made  after  starting  equals  the  square  of  the  time  (in 
minutes)  after  starting.  Deduce  expressions  for  the  angular 
velocity  and  the  acceleration  at  any  time. 

Solution:  The  law  of  the  motion,  if  d  and  t  denote  the  num- 
ber of  turns  and  the  time,  is  ^  =  ^^;  hence 

dd/dt  =  2t  =  aj  (turns-per-min.) , 

and  dW/df  =  2  =  a  (turns-per-min.-per-min.). 

5.  Compute  the  velocity  and  the  acceleration  of  a  point  on 
the  rim  of  a  wheel  whose  diameter  is  six  feet,  if  its  angular 
velocity  is  4  rev.-per-sec. 

Solution:  w  =  4 X6. 283  =  25.13  rad./sec.     u; '--^X 

According  to  eq.  (i),  v  =  ^w  =^$.4  ft. /sec,  and   '-  ^ 

"eq.(2),  a„  =  'z;V3  =  i895ft./sec.2 

Since  the  angular  velocity  is  constant,  the  angular  acceleration 
is  zero;  hence,  see  eqs.  (2)  and  (3),  a^  =  o  and  a  =  a„.  The  direc- 
tion of  a  is  from  the  point  on  the  rim  to  the  centre  of  the  wheel. 

§  III.     Plane  Motion. 

217.  Plane  Motion  Defined. — A  plane  motion  is  one  in  which 
each  ppint  of  the  moving  body  remains  at  a  constant  distance 
from  a  fixed  plane.  The  fixed  plane  (or  any  plane  parallel  to  it) 
is  called  the  plane  of  the  motion. 

The  wheel  of  a  car  running  on  a  straight  track  has  plane 
motion;  so  also  has  a  book  sliding  about  on  a  table.  A  trans- 
lation may  or  may  not  be  a  plane  motion  (see  illustrations,  art. 
205),  but  a  rotation  is  always  a  plane  motion. 

As  in  a  rotation,  all  points  on  any  line  of  the  moving  body 


^oS  MOTION  OF  A  RIGID  BODY,  [Chap.  IX.; 

which  is  perpendicular  to  the  plane  of  the  motion  move  alike, 
hence  the  motion  of  the  projection^of  the  line  on  the  plane  of 
the  motion  represents  that  of  all  the  points,  and  the  motion  of' 
the  body  Itself  is  completely  represented  by  that  of  its  projec- 
tion on  the  plane  of  the  motion, 

218.  Angular  Displacement. — By  angular  displacement  of  a 
body  whose  rriotion  is  plane  is  meant  (as  in  rotations)  the  angle 
described  by  any  line  of  the  body  which  is  parallel  to  the^ 
plane  of  the  motion.  Obviously  all  such  lines  describe  equah 
angles  in  the  same  time  interval.  As  in  rotations  also,  displace- 
ments are  regarded  as  positive  or  negative  according  as  they 
are  due  to  counter-clockwise  or  clockwise  turning  of  the  body. 

Let  the  irregular  outline  (fig.  174)  represent  the  projection 
of  the  moving  body  on  the  plane  of  the. 
motion,  AB  a  fixed  line  of  the  projection,: 
OX  a  fixed  reference  line,  and  let  d  denote 
the  angle  XOA ,  it  being  regarded  as  posi- 
tive or  negative  according  as  OX,  when, 
turned  about  0  toward  AB,  turns  counter-^ 
clockwise  or  clockwise.     If  6^  and  ^2 denote  ] 
initial  and  final  values  of  0  corresponding  to  ■ 
Fig.  174.  any  motion  of  the  body,  then  the 

angular  displacement  =  62  —  6^  =  Ad. 

• 

219.  Angular  Velocity  and  Angular  Acceleration. — If  a  body 

has  a  plane  motion,  its  angular  velocity  is  the  time-rate  at  which  ' 
its  displacement  occurs,  and  its  angular  acceleration  is  the  time-  ' 
rate  at  which  its  angular  velocity  changes.  1 

These  definitions  are  precisely  similar  to  those  of  the  angular  j 
velocity  and  acceleration  of  a  rotation  (arts.  210  and  213) ;  hence  \ 
the  expressions,  units,  and  rules  of  signs  given  in  those  articles  j 
and  the  following  hold  for  any  plane  motion.  Rewriting  the  ; 
expressions,  \ 

o;  =  d^/dt,     and     a  =  dw/dt  =  d^^/dt^  \ 

CO  and  a  denoting  angular  velocity  and  acceleration  of  the  mov-  : 
ing  body  respectively. 


§  111.]  PLANE  MOTION.  i^OQ 

EXAMPLES. 

I.  Determine  the  angular  velocity  and  acceleration  of  the 
connecting-rod  of  a  steam-engine,  assuming  the  angular  veloc- 
ity of  the  crank  to  be  constant. 

Solution:  Let  OP  and  CP  (fig.  175)  be  the  crank  and  con- 
necting-rod respectively,  their  lengths  being  c  and  r,  and  let  the 


Fig.  175. 

angles  XOP  and  XCP  be  denoted  by  6  and  ^  (measured  from 
the  horizontal  line  and  counter-clockwise  positive) ;  then  for  all 
positions 

/^Gi  -    r  sin  ^  =  c  sin  ^,-   or     ^  =  sin~M -sin  ^  J. 

Let  CO  and  a  denote  the  angular  velocity  and  acceleration  of  the 
rod  respectively,  and  (Oc  the  angular  velocity  of  the  crank.  Then, 
since 

(jj  =  d^/dt,     a  =  d^(f)/df,     (Oc  =  dd/dt,     and     d^d/dt^^o,    Coj/>i 

cos  d  ,   . 

(rV^^  — sm^  d)^ 

and  a=  —.0/2 •   2  /)n?  ^c  ♦ (2) 

2.  Take  r/c  equal  to  4  and  compute  the  values  of  the  coeffi- 
cients of  ojc  and  oj^  in  eqs.  (i)  and  (2)  of  the  preceding  solution, 
when  ^  =  0°,  30°,  60°,  90°,  120°,  150°,  and  180°.  Plot  those 
values,  thus  showing  how  oj  and  a  vary  during  a  stroke. 

220.  Velocity  and  Acceleration  of  any  Point  of  the  Body. — 
Let  P  and  P'  (fig.  176)  be  two  points  of  the  moving  body  and  0 
a  point  without,  the  three  being  in  a  plane  parallel  to  that  of  the 
motion.  According  to  art.  196  the  velocity  of  P  relative  to  0 
equals  the  vector  sum  of  the  velocities  of  P  relative  to  P'  and  P' 
relative  to  0.  Let  the  last  velocity  be  v'  directed  as  shown,  and 
let  r  denote  the  distance  PP'  and  oj  the  angular  velocity  of  the 


2IO 


MOTION  OF  A  RIGID  BODY. 


[Chap.  IX. 


body  at  the  instant  considered.  Since  the  motion  of  P  relative 
to  P'  is  circular,  the  velocity  of  P  relative  to  P'  equals  rco  (art. 
216),  and  its  direction  is  perpendicular  to  PP'  as  shown.  The 
velocity  of  P  relative  to  O  then  is  the  vector  sum  of  v'  and  roj. 


(a)  (6) 

Fig.  176. 

Similarly,  the  acceleration  of  P  relative  to  O  equals  the  vec- 
tor sum  of  the  accelerations  of  P  relative  to  P'  and  P'  relative 
to  0.  Let  the  last  acceleration  be  a'  directed  as  shown,  and  let 
a  denote  the  angular  acceleration  of  the  body  at  the  instant  con- 
sidered. The  path  of  P  relative  to  P'  being  a  circle,  the  tan- 
gential and  normal  components  of  the  acceleration  of  P  relative 
to  P'  are  respectively  ra  and  roJ^  (art.  216)  directed  as  shown, 
and  the  acceleration  of  P  relative  to  O  is  the  vector  sum  of  a', 
ra,  and  roj^. 

Proposition. — The  components  of  the  velocities  of  any  two 
points  of  a  body  having  a  plane  motion  *  along  the  line  join- 
ing them  are  equal  and  agree  in  sense. 

Proof:  Let  P  and  P'  (fig.  176a)  be  two  points  in  the  plane  of 
the  motion.  Since  the  velocity  of  P  (v)  is  the  resultant  of  rco 
and  ^-'and  rco  is  perpendicular  to  PP\  the  component  of  v  along 
PP'  is  the  same  as  that  of  v^  along  that  line ;  but  v^  is  the  veloc- 
ity of  P',  hence,  etc. 

It  will  be  noticed  that  the  proof  is  not  general,  the  points 
being  in  the  plane  of  the  motion;  this  vj  the  case  to  which  the 
proposition  is  applied  later. 


*  Really  true  i,n  any  motion  of  a  rigid  body. 


i      §111.]  PLANE  MOTION.  211 

EXAMPLES. 

1 .  A  wheel  rolls  uniformly  and  makes  one  turn  every  second. 
What  is  its  angular  velocity? 

Solution:  Any  line  of  the  body  parallel  to  the  plane  of  the 
wheel  describes  an  angle  of  360°  each  second;  hence  the  angular 
velocity  is  360  deg.-per-sec. 

2.  Determine  the  velocity  of  any  point  on  the  rim  of  a  wheel 
of  radius  r  whose  angular  velocity  is  (o  rad. -per-unit  time. 

Solution:  We  use  the  foregoing  principles,  choosing  the 
centre  of  the  wheel  as  P'.     Let  d  denote  the  angle  described  by 

the  radius  PP'  after  any  origin  of  time ;  .   p 

then  the  distance  (5)  travelled  by  P'  in  '^^^'^^^^^'-^^r^S^^-A  ^^^ 

that   time   is   given   by   5  =  rd.     Hence         /^        \  ^^'^^^ 

ds/dt==r  dd/dt,  or  the  velocity  of  P'  at      /  rw  ^/A 

any  instant  equals  r  times  the  angular     /  ^^^  \ 

velocity  of  the  wheel  at  that  instant  (see     I       "^rw 

fig-  177)-     Relative  to  P',  the  selected      \ 

point  P  on  the  rim  describes  a  circle  and       \ 

the  velocity  of  that  point  relative  to  P'  ^^^ 

is  roj  (art.  216),  its  direction  being  that         j^mmmMimw 

of  the  tangent  to  the  circle  at  P.     Hence  Fig.  177. 

the  absolute  velocity  of  P  (v)  is  represented  by  the  diagonal  Pp 

of  the  parallelogram  on  the  two  velocities  roj. 

3.  From  the  result  of  the  preceding  solution,  determine  the 
velocities  of  the  highest  and  lowest  points  on  the  rim  of  the 
wheel.    '    ^^^     -->  ^ 

221.  Plane  Motion  Regarded  as  a  Combined  Translation  and 
Rotation. — Imagine  the  velocity  of  each  point,  P^,  P^,  etc.,  of 
the  moving  body  to  be  resolved  into  two  components,  one  of 
which  is  the  same  as  the  velocity  of  any  particular  point  P'  of 
the  body  (fig.  178a).  It  follows  from  the  preceding  article  that 
the  other  component  is  perpendicular  to  the  line  joining  the 
point  with  P'  and  is  equal  to  the  product  of  the  length  of  the 
line  and  the  angular  velocity  of  the  body. 

Also  imagine  the  acceleration  of  each  point  resolved  into  two 
components,  one  of  which  is  the  same  as  the  acceleration  of  P' 
(fig.  1786).     It  follows  from  the  preceding  article  that  the  other 


212 


MOTION  OF  A  RIGID  BODY. 


[Chap.  IX. 


component  can  be  resolved  into  two  components  directed  along 
and  perpendicular  to  the  line  joiniijg  the  point  with  P\  they 
being  equal  to  the  products  of  the  length  of  the  line  and  the 
square  of  the  angular  velocity  and  angular  acceleration  respec- 
tively. 


Fig.  178.  , 

Now  if  the  points  of  the  body  had  the  first^sets  of  component 
velocities  and  accelerations  only,  the  motion  of  the  body  would 
be  a  translation,  the  velocity  and  acceleration  of  which  would  be 
like  those  of  the  chosen  point  P' .  And  if  the  particles  had  the 
second  sets  of  component  velocities  and  accelerations  only,  the 
motion  of  the  body  would  be  a  rotation  about  the  line  through 
the  chosen  point  and  perpendicular  to  the  plane  of  the  motion, 
the  angular  velocity  and  acceleration  of  the  rotation  equal- 
ing respectively  the  angular  velocity  and  acceleration  of  the 
actual  motion.  The  two  motions  are  hence  regarded  as  compo- 
nents of  the  actual  motion. 

222.  Instantaneous  Axis  (of  no  Velocity). — Proposition. — If 

a  body  has  a  plane  motion  which 
is  not  translatory,  then  at  each 
instant  there  is  a  line  in  it  or  in 
its  extension  all  points  of  which 
have  no  velocity. 

Proof:  Let P and P'  (fig.  179) 
be  any  two  points  in  the  plane 
of  the  motion,  and    draw  lines 
PO  and  P'O  perpendicular  to  the  directions  of  the  velocities  of 


§  III.  ]  PLANE  MO  TION.  2 1 3 

P  and  P'  respectively.  The  velocities  of  points  on  PO  and  P'O 
have  no  components  along  those  lines  respectively  (prop.,  art. 
220),  and  since  O  is  on  both  lines  its  velocity  has  no  components 
along  PO  and  P'O,  and  hence  that  velocity  must  be  zero.  All 
points  on  the  line  through  O  perpendicular  to  the  plane  of  the 
motion  have  the  same  velocity  as  that  of  0,  i.e.,  zero. 

If  the  motion  is  translatory,  the  velocities  of  P  and  P'  have 
the  same  direction  and  the  point  O  is  *'  at  infinity,"  but  no  points 
of  a  finite  extension  of  the  body  have  a  zero  velocity. 

Definitions. — The  line  of  a  moving  body,  all  points  of  which 
have  at  a  certain  instant  no  velocity,  is  called  the  instantaneous 
axis  of  the  motion  at  that  instant.  The  intersection  of  the 
instantaneous  axis  with  the  plane  of  the  motion  is  called  the 
instantaneous  centre. 

In  general  the  instantaneous  centre  moves  about  in  the  body 
and  in  space.  Its  path  in  the  body  (i.e.,  the  path  relative  to 
axes  fixed  in  the  body)  is  called  body  centrode^  and  its  path  in 
space  (i.e.,  that  relative  to  axes  fixed  in  space,  or  in  the  earth) 
is  called  space  centrode. 

It  follows  from  the  solution  of  exs.  2  and  3,  art.  220,  that  the 
velocity  of  the  lowest  point  of  a  rolling  wheel  is  zero ;  hence  that 
point  is  the  instantaneous  centre.  The  body  centrode  is  the 
circumference  of  the  wheel,  and  the  space  centrode  is  the  line  on 
the  plane  surface  along  which  the  wheel  rolls. 

223.  Instantaneous  Rotation. — Let  P  and  Q  denote  two 
points  of  a  body  in  a  plane  of  motion,  the  latter  being  such 
that  at  some  instant  during  the  motion  it  is  an  instantaneous 
centre.  At  all  times  the  motion  of  P  relative  to  Q  is  circular 
and  its  velocity  relative  to  Q  at  any  instant  equals  the  product 
of  the  distance  between  P  and  Q  (r)  and  the  angular  velocity  of 
the  body  at  that  instant. 

Consider  now  the  state  of  the  motion  when  Q  is  the  instan- 
taneous centre,  calling  the  angular  velocity  of  the  body  at  that 
instant  co.  Since  the  absolute  velocity  of  Q  is  zero,  the  absolute 
velocity  of  P  (v)  equals  its  velocity  relative  to  Q,  or 

Y  =  TOJ (l) 

Hence  the  absolute  velocity  of  any  point  of  a  mov- 


3^W^^^ 


214 


MOTION  OF  A  RIGID  BODY. 


lChap.  IX. 


ing  body  at  any  instant  equals  the  product  of  the 
distance  of  the  point  from  the»  instantaneous  axis 
and  the  angular  velocity  of  the  body  at  that  in- 
stant. 
Since  in  a  rotation  about  a  fixed  axis  the  velocity  of  any 
point  of  the  body  is  also  proportional  to  its  distance  from  the 
axis  of  rotation,  the  state  of  a  plane  motion  at  any  instant  is 
described  as  an  instantaneous  rotation  about  the  instantaneous 
axis. 

EXAMPLES. 

1.  Show  how  to  find  the  instantaneous  centre  of  the  connect- 
ing-rod of  a  steam-engine  in  any  position. 

Solution:  The  directions  of  the  velocities  of  the  ends  of  the 
rod  are  known  for  all  positions;  that  of  C  (fig.  i8o)  is  DC  and 
that  of  P  is  the  same  as  that  of  the  tangent  to  the  crank-pin 
circle  at  P.  Hence  to  find  the  instantaneous  centre,  draw  per- 
pendiculars to  the  directions  of  these  velocities  atil  and  P  re- 
spectively ;  their  intersection  is  the  instantaneous  centre  of  the 
rod  for  the  position  represented. 

2.  Where  is  the  instantaneous  centre  when  the  crank  is  hor- 
izontal? When  vertical?  What  can  you  say  about  the  state 
of  the  motion  of  the  rod  in  these  cases  ? 

3.  Show  how  to  find  the  angular  velocity  of  the  connecting- 


FiG.  180. 


rod  and  the  velocity  of  the  cross-head  C  (fig.  180)  in  terms  of 
the  velocity  of  the  crank-pin  P  for  any  position. 


§111.]  PLANE  MOTION.  215 

Solution:  Let  c  denote  the  length  of  the  crank; 

v^,     **         '*    velocity  of  the  crank-pin ; 
V2,     "         **         **  *'    "    cross-head; 

(D,      "         "    angular  velocity  of  the  connecting - 
rod. 

According  to  eq.  (i),  (o  =  vJlP  and 

If  it  is  desired  to  draw  a  velocity-space  curve  for  the  motion 
of  the  cross-head  (as  the  dotted  curve  of  the  figure)  when  the 
crank-pin  velocity  is  constant,  the  following  simple  construc- 
tions may  be  employed :  Draw  a  vertical  diameter  of  the  crank- 
pin  circle  and  mark  its  intersection  with  the  connecting-rod 
(extended  if  necessary)  C'\  then  to  the  same  scale  by  which  OP 
represents  the  crank-pin  velocity  OC  represents  the  cross-head 
velocity.     For,  from  the  figure, 

IC/rP  =  OC'/c,     hence     vJv,=^OC'/c. 

4.  A  sheet  of  paper  is  caused  to  slide  on  a  draughting- board 
so  that  two  points  (P  and  Q)  of  the  paper  move  along  two  lines 
{OX  and  OY)  on  the  board.  Show  how  to  find  the  instanta- 
neous centre  of  the  motion  for  any  position  of  the  paper. 

5.  The  velocity  of  one  point  of  the  paper  of  the  preceding 
example  being  given,  show  how  to  find  the  velocity  of  any  other 
point. 

EXERCISE. 

Take  a  sheet  of  paper  and  move  it  as  described  in  ex.  4,  the 
lines  OX  and  OY  being  taken  at  right  angles  to  each  other,  and 
determine  the  instantaneous  centre  for  several  positions  of  the 
paper.  As  each  instantaneous  centre  is  determined,  mark  it  by 
pricking  a  hole  through  the  paper  and  into  the  board,  and  join 
the  holes  in  the  paper  and  those  in  the  board  by  smooth 
curves.  These  curves  are  the  body  and  space  centrodes 
respectively. 

Now  cut  the  paper  along  the  centrode  and  replace  it  on  the 


2i6  MOTION  OF  A  RIGID  BODY,  [Chap.  IX. 

board  in  one  of  its  positions  (P  and  Q  falling  on  OX  and  OY 
respectively).  Then  move  the  paper^so  that  the  body  centrode 
rolls  on  the  space  centrode.  If  carefully  done  (a  template  fitted 
to  the  space  centrode  helps  to  get  the  rolling  motion),  it  will  be 
noticed  that  P  and  Q  move  along  OX  and  0Y\  hence  the  actual 
motion  is  a  rolling  of  the  body  centrode  over  the  space  centrode. 
It  can  be  shown  that  any  plane  motion  is  equivalent  to  such  a 
rolling. 


KINETICS, 


CHAPTER   X. 

MOTION  OF  A  PARTICLE  (RESUMED)  AND  OF  A  SYSTEM  OF 

PARTICLES. 

§  I.     Mass  and  Mass-Centre. 

224.  Quantity  of  Matter. — It  must  be  confessed  that  the 
usual  definition  of  mass  (art.  12)  needs  explanation.  The  ques- 
tion at  once  arises,  How  shall  matter  be  measured?  If  this  is 
answered,  then  the  meaning  of  "quantity  of  matter"  is  clear. 

"As  long  as  we  have  to  do  with  bodies  of  the  same  exact  kind, 
there  is  no  difficulty  in  understanding  how  the  quantity  of  mat- 
ter is  to  be  measured.  If  equal  quantities  of  the  substance 
produce  equal  effects  of  any  kind,  we  may  employ  these  effects 
as  measures  of  the  quantity  of  the  substance. 

"For  instance,  if  we  are  dealing  with  sulphuric  acid  of  uni- 
form strength,  we  may  estimate  the  quantity  of  a  given  portion 
of  it  in  several  ways.  We  may  weigh  it,  we  may  pour  it  into  a 
graduated  vessel  and  so  measure  its  volume,  or  we  may  ascertain 
how  much  of  a  standard  solution  of  potash  it  will  neutralize. 

"We  might  use  the  same  methods  to  estimate  a  quantity  of 
nitric  acid  if  we  were  dealing  only  with  nitric  acid;  but  if  we 
wished  to  compare  a  quantity  of  nitric  acid  with  a  quantity  of 
sulphuric  acid  we  should  obtain  different  results  by  weighing,  by 
measuring,  and  by  testing  with  an  alkaline  solution."  * 

Now  these  methods  are  not  equally  appropriate,  and  indeed 
that  of  titration  cannot  be  applied  to  all  bodies,  that  of  measur- 
ing would  lead  us  to  say  that  the  amount  of  gas  in  a  tight  rubber 
bag  could  be  changed  by  simply  squeezing  (changing  the  volume 

*  Quoted  from  Maxwell's  "Matter  and  Motion." 

217 


2l8 


MOTION  OF  A  PARTICLE. 


[Chap.  X. 


of  the  bag),  which  is  absurd.     It  will  be  shown  that  the  method 
of  weighing  is  an  appropriate  one.      , 

Any  appropriate  method  must  be  based  on  a  common  prop-     i 
erty  of  matter.     Now  all  matter  is  inert,  i.e.,  force  must  be 
applied  to  any  body  to  change  its  velocity,  and  this  property 
(inertia)  is  the  basis  of  the  fundamental  method  of  determining     i 
quantity  of  matter.     Not  only  is  this  method  (explained  in  de- 
tail later)  employed  in  mechanics,  but  also  sometimes  in  ordi-     \ 
nary  affairs.     Thus,  suppose  that  we  wish  to  ascertain  whether 
a  barrel  lying  upon  a  floor  is  full  or  empty ;  we  push  it  and  con- 
clude that  it  is  full  or  empty  according  as  a  large  or  small  force 
is  required  to  roll  it,  i.e.,  to  "overcome  the  inertia." 

Along  this  line  we  can  determine  the  relative  amounts  of  i 
matter  in  two  bodies  if  we  have  a  means  of  measuring  forces.  ^ 
Thus,  calling  the  two  bodies  A  and  B,  place  each  upon  a  light  ; 
and  easy-running  carriage  (fig.  1 8 1 )  and  connect  these  by  means    ] 


Fig.  i8i. 

of  cords  to  spring-balances  which  rest  upon  and  are  fastened  to 
a  third  carriage  as  shown.  If  this  last  carriage  is  pulled  to  the 
right,  the  others  follow  and  the  spring-balances  measure  the  pulls 
required  to  move  the  smaller  carriages  and  their  loads.  Clearly 
it  will  be  in  accord  with  the  crude  test  applied  to  the  barrels  to 
say  that  the  quantities  of  matter  in  A  and  B  (neglecting  that  in 
the  carriages)  are  as  the  forces  applied  to  them,  as  read  from 
the  spring-balances.  It  will  also  be  in  accord  with  a  comparison 
of  quantities  of  matter  in  bodies  of  the  same  kind  by  the  method 
of  volumes ;  for  suppose  that  A  and  B  are  of  the  same  kind  and 
that  the  volume  of  ^4  is  w  times  that  of  J5,  we  would  say  at  once 
that  there  is  n  times  as  much  matter  in  ^  as  in  5.     The  same 


§  I.]  MyiSS  AND  MASS-CENTRE.  219 

ratio  would  be  arrived  at  by  the  inertia  method,  i.e.,  from  the 
readings  of  the  spring-balances. 

We  now  make  definite  our  notions  about  quantity  of  matter 
as  just  expressed  by  means  of  the  following 

Definition. — The  quantities  of  matter  in  bodies  are  propor- 
tional to  the  forces  required  to  give  them  equal  accelerations. 

If  we  should  adopt  as  a  standard,  or  unit,  ^the  amount  of 
matter  in  any  particular  body,  we  could  determine  the  quantity 
in  any  other  body  in  terms  of  this  unit  (ideally  at  least)  by  plac- 
ing it  and  the  standard  body  on  the  two  smaller  carriages  and 
then  measuring  the  forces  required  to  give  them  the  equal  accel- 
erations. If  the  force  on  the  body  in  question  is  n  times  that 
on  the  standard,  the  quantity  of  matter  in  the  body  would  be  n. 

Of  course  this  scheme  of  measurement  is  not  put  forth  as  a 
practical  one,  but  rather  as  a  help  to  understand  the  meaning  of 
mass.  The  practical  method  of  measuring  quantity  of  matter  is 
by  weighing,  which  is  (as  explained  in  art.  226)  precisely  similar 
in  principle  to  that  just  described. 

225.  Mass. — Definition  (repeated  from  art.  12). — By  mass  of 
a  body  is  meant  the  quantity  of  matter  in  it.  The  word  is 
merely  an  abbreviation  for  "  quantity  of  matter." 

Units  of  Mass. — There  are  many  units  of  mass  in  use;  they 
may  be  grouped  into  two  classes: 

(a)  Absolute  Units;  so  called  to  express  the  fact  that  their 
magnitudes  are  independent  of  locality.  Two  of  these,  the 
pound  and  the  kilogram,  are  described  in  art.  12.  Their  rela- 
tion is  as  follows: 

I  pound  =  0.4536  kilograms, 
or  I  kilogram  =  2.205  pounds. 

(6)  Gravitation  Units;  so  called  because  their  magnitudes 
vary  with  locality  precisely  as  acceleration  due  to  gravity  varies. 
Two  of  these  units  are  described  in  art.  233. 

226.  Practical  Determination  of  Mass. — As  is  well  known, 
bodies  falling  at  the  same  place  in  vacuum  move  with  equal 
accelerations,  i.e.,  the  forces  of  gravity  upon  bodies  (their  weights) 
at  the  same  place  accelerate  them  equally.  It  follows  from  this 
fact  and  the  definitions  of  arts.  224  and  225  that 


220  MOTION  OF  A  PARTICLE.  [Chap.  X.  \ 

the  masses  of  bodies  are  proportional 
to  their  weights  at  the  same  place. 
Hence  to  determine  the  mass  of  a  body,  i.e.,  to  determine  the  : 
ratio  of  its  mass  to  that  of  a  standard  (pound  for  example) ,  we  , 
determine  the  ratio  of  the  weights  of  the  body  and  the  standard.  : 
If  this  latter  ratio  is  n,  then  the  former  is  n  also,  and  the  mass  ' 
of  the  body  is  n  times  that  of  the  standard  (or  n  pounds).  J 

227.  Moment  of  Mass. — The  product  of  the  mass  of  a  particle  ; 
and  its  ordinate  with  respect  to  any  plane  is  called  the  moment  ■ 
of  the  mass  of  the  particle  with  respect  to  that  plane.  An  ordi-  i 
nate  is  regarded  as  positive  or  negative  according  as  the  parti-  ^ 
cle  is  on  one  side  of  the  plane  or  the  other ;  hence  a  moment  has  \ 
the  same  sign  as  the  corresponding  ordinate.  \ 

By  moment  of  the  mass  of  a  system  of  particles  is  meant  the  \ 
algebraic  sum  of  the  moments  of  the  masses  of  its  particles.  1 
Thus,  let  the  particles  of  a  system  be  referred  to  a  set  of  rect-  ; 
angular  axes,  and  denote  the  coordinates  of  the  particles  by  ; 
(Xj,  ^1,  z^,  (x^,  ^2*  ^2)'  ^"tc,  and  their  masses  by  m^,  m^,  etc.  Then  ; 
the  moments  of  the  mass  of  the  system  with  respect  to  the  y-z,  ] 
z-x,  and  x-y  planes  are  respectively 

m^x^-\-m2X2-\-  ,  .  .  ^Imx;  i 

Wi>/i+W2>'2+  •  •  •  =^niy',  1 

miZi+m2Z2  +  .  .  .  =  Imz,  '  \ 

'1 

228.  Mass-Centre  Defined. — It  is  obvious  that  the  mass  of  a  \ 
system  may  be  multiplied  by  some  distance  (positive  or  nega-  \ 
tive)  such  that  the  product  equals  the  moment  of  the  mass  with  \ 
respect  to  a  given  plane.  Thus,  let  M  be  the  mass  of  a  system  j 
and  x,y,  and  z  such  multipliers  that  \ 

Mx  =  Imx,     My  =  Jmy,     Mz  =  Imz.    .     .     .     (i)  ■ 

The  point  whose  coordinates  are  x,  y,  and  z  is  called  the  mass-  ; 
centre  of  the  system.  The  formulas  for  the  coordinates  of  the  '■ 
mass-centre  are  therefore 

,     Jnix      ,  _  Jmy      -Jniz  .  .  ^ 

^~    M   *     y~    M  '     ^~~W ^^^  i 

1 

i 

I 

\ 
J 


:    §11.]  MOTION  OF  A   PARTICLE.  221 

229.  Relation  Between  Mass-Centre  and  Centre  of  Gravity. — 

Since  the  masses  of  the  particles  of  a  system  are  proportional 
to  their  weights,  we  may  substitute  for  the  mass  terms  in  the 
equations  (2)  the  corresponding  weights,  i.e., 

_  _  Iw'x      _  _  Iwy      _  _  Iwz 

^^~W'    ^^1V~'    ^~~W' 

w  denoting  the  weight  of  any  particle  whose  coordinates  are  x,  y, 
and  z,  and  W  the  weight  of  the  system.  But  these  values  of  x, 
y,  and  z  are  identical  with  those  for  the  x,  y,  and  z  coordinates 
of  the  centre  of  gravity  of  the  system  (see  art.  64) ;  hence  the 
mass-centre  and  centre  of  gravity  of  a  system  of  particles  (as 
herein  defined)  are  coincident. 


§  II.     Motion  of  a  Particle. 

230.  Laws  of  Motion. — (i)  When  no  force  is  exerted  upon  a 
particle  it  remains  at  rest  or  continues  to  move  uniformly  in  a 
straight  line. 

(2)  When  a  force  is  exerted  upon  a  particle  it  is  accelerated; 
the  direction  of  the  acceleration  is  the  same  as  that  of  the  force, 
and  its  magnitude  is  proportional  to  the  force  directly  and  to  the 
mass  of  the  particle  inversely.        £^  ^  ^^ 

(3)  When  one  particle  exerts  a  force  upon  another  the  latter 
exerts  one  on  the  former,  and  the  two  forces  are  equal,  coUinear, 
and  opposite.* 

These  laws  are  inductions  from  observation  and  experiment, 
made  not  of  course  on  particles,  but  on  bodies  of  ordinary  size. 
We  do  not  attempt  a  full  discussion  of  the  experience  leading  to 
the  laws,  but  limit  ourselves  to  a  brief  statement.  Really  the 
best  evidence  of  the  correctness  of  the  laws  are  the  many  agree- 
ments noted  between  observed  results  and  those  calculated  from 
the  laws,  and  the  fact  that  the  laws  are  not  known  to  be  in  dis- 
agreement with  any  phenomenon. 

(i)  We  know  of  no  body  which  is  free  from  the  influence  of 

*  These  are  essentially  Newton's  Laws  of  Motion.  The  form  of  state- 
ment here  given,  however,  differs  from  that  in  which  they  were  originally 
announced  (1687), 


Ji22  MOTION  OF  A  PARTICLE.  [Chap.  X, 

all  others,  i.e.,  a  body  not  acted  upon  by  force,  and  so  no  direct 
observation  or  experiment  leads  to  this  law.  But  all  have  per- 
haps noticed  that  a  small  body,  if  projected  along  a  sheet  of  ice, 
moves  in  a  straight  path  and  continues  to  move  for  a  consider- 
able period;  also  that  the  smoother  the  ice  the  longer  will  the 
body  move,  i.e.,  the  smaller  is  its  retardation  or  the  more  nearly 
is  its  motion  uniform.  The  retardation  is  rightly  ascribed  to  the 
frictional  resistance  offered  by  the  ice ;  and  it  is  a  fair  inference 
that  if  that  resistance  were  zero,  the  retardation  would  also  be 
zero  and  the  motion  would  be  uniform. 

(2)  The  second  law  may  be  roughly  verified  by  the  means  of 
the  apparatus  represented  in  fig.  181.  Thus  place  a  body  whose 
mass  is  known  on  one  of  the  small  carriages  and  another  on  the 
hook  C.  Then  let  the  system  move,  measure  the  acceleration  of 
A  and  the  pull  recorded  by  the  spring-balance.  If  this  is  re- 
peated with  other  bodies  on  (7,  it  will  be  found  that  the  acceler- 
ations are  roughly  proportional  to  the  pulls.  Also  place  bodies 
of  unequal  mass  successively  on  A  and  cause  the  carriages  to 
roll  so  that  the  spring-balance  reading  is  the  same  for  the  differ- 
ent bodies,  measuring  the  acceleration  of  each  motion.  It  will 
be  found  that  the  accelerations  are  roughly  inversely  propor- 
tional to  the  masses  of  the  bodies.  The  lack  of  exact  propor- 
tionality may  properly  be  ascribed  to  the  neglect  of  friction  and 
the  mass  of  the  carriage  A,  and  to  unavoidable  experimental 
errors  involved  in  such  an  apparatus. 

(3)  It  is  well  known  that  a  magnet  exerts  a  force  on  a  piece 
of  soft  iron  in  its  proximity ;  that  the  iron  also  exerts  a  force  on 
the  magnet,  may  be  proven  as  follows:  Place  the  magnet  and 
piece  of  iron  in  small  vessels ;  then  float  them  so  loaded  in  water. 
If  they  are  not  too  far  apart  they  will  be  observed  to  move 
toward  each  other;  hence  not  only  does  the  magnet  attract  the 
soft  iron,  but  the  iron  attracts  (exerts  a  force)  on  the  magnet. 

If  two  spring-balances  be  laid  on  a  table,  the  hook  of  one 
engaging  the  hook*  of  the  other,  and  then  they  be  pulled  apart 
by  their  rings,  each  will  register  the  pull  exerted  upon  it  by  the 
other.  If  the  balances  are  accurate,  it  will  be  found  that  the 
amounts  registered  are  equal;  hence  the  forces  exerted  by  the 
hooks  on  each  other  are  equal.        /^  ^  f^ 


§  II.]  MOTION  OF  A  PARTICLE,  223 

231.  Quantitative  Expression  of  the  Second  Law  of  Motion. — 

Let  m'  and  m"  denote  the  masses  of  two  particles  and  a'  and  a" 
the  accelerations  given  them  by  two  forces  F'and  F"  respectively. 
The  second  law  of  motion  asserts  that 

a' :  a" :  :  F' /m' :  F" /m" ,     or     F'/m'a'  =  F"lm"a". 

It  follows  from  this  equation  that  the  ratio  of  the  force  acting 
on  a  particle  to  the  product  of  the  mass  and  the  acceleration  cf 
the  particle  produced  by  the  force  is  constant,  i.e.,  in  the  form 
of  an  equation, 

F/ma  =  k,     or     F  =  kma, (i) 

h  being  the  value  of  the  constant  ratio. 

The  numerical  value  of  k  depends  upon  the  units  used  for 
expressing  value  of  force,  mass,  and  acceleration.  It  is  con- 
venient, but  not  necessary,  so  to  choose  these  units  that  k 
becomes  i.  A  system  of  units  so  chosen  is  called  a  kinetic 
system. 

232.  Kinetic  System  of  Units. — It  follows  from  the  last  equa- 
tion that  in  any  kinetic  system  the  unit  force  acting  upon  the 
unit  mass  produces  unit  acceleration  (i.e.,  unit  velocity  in  unit 
time).  Evidently  any  three  of  the  units  involved  in  a  kinetic 
system  (units  of  force,  mass,  length,  and  time)  may  be  chosen 
arbitrarily,  but  the  fourth  must  be  such  as  to  satisfy  the  re- 
quirement just  stated.  The  unit  of  force  or  of  mass  is  the  one 
selected  as  the  derived  or  fourth  unit.  There  are  two  classes  of 
kinetic  systems. 

(a)  Absolute  Kinetic  Systems. — Units  of  length,  mass,  and 
time  are  arbitrarily  selected,  the  unit  of  force  being  derived. 

Centimetre-Gram-Second  (cg.s.)  System.  —  The  units  of 
length,  mass,  and  time  are  the  centimetre,  the  gram  (one-thou- 
sandth of  a  kilogram,  see  art.  12),  and  the  second  respectively. 
The  corresponding  unit  of  force,  i.e.,  the  force  which  acting  on 
a  gram  mass  for  one  second  gives  it  a  velocity  of  one  centimetre 
per  second,  is  called  a  dyne.  This  is  the  system  now  univer- 
sally used  in  scientific  work  and  literature. 

Foot-Pound-Second  (f.p.s.)  System. — The  units  of  length, 
mass,    and   time    are   the   foot,  the   pound   (see  art.   12),  and 


224  MOTION  OF  A  PARTICLE.  [Chap.  X. 

the  second  respectively.  The  corresponding  unit  of  force,  i.e. ,  the 
force  which  acting  on  a  pound  mass  for  one  second  gives  it  a 
velocity  of  one  foot  per  second,  is  called  a  poundal.  This  sys- 
tem has  never  met  with  favor;  accordingly,  it  is  not  used  herein, 
but  is  mentioned  and  explained  because  of  the  relations  which 
it  bears  to  other  important  systems. 

(6)  Gravitation  Kinetic  Systems  (also  called  Engineers'  Sys- 
tems).— Units  of  length,  force,  and  time  are  arbitrarily  selected 
and  the  unit  of  mass  is  derived.  The  units  of  force  in  these 
systems  are  weights  of  absolute  units  of  mass,  i.e.,  they  depend 
upon  the  force  of  gravity  and  are  gravitation  units  (see  art.  13) 
The  units  of  length  and  time  being  independent  of  place,  the  units 
of  mass  in  these  systems  must  vary  just  as  the  units  of  force 
vary,  and  are  hence  properly  called  gravitation  units  of  mass. 

Foot-Pound  (force)  -Second  System. — The  units  of  length, 
force,  and  time  are  the  foot,  the  pound  (see  art,  13),  and  the 
second  respectively.  The  corresponding  unit  of  mass,  i.e.,  one  in 
which  a  pound  force  would  produce  in  one  second  a  velocity 
of  one  foot-per-second,  is  called  herein  a  gee  pound. ^ 

Metre-Kilogram  (force)  -Second  System. — The  units  of  length, 
force,  and  time  are  the  metre,  the  kilogram  (see  art.  13),  and  the 
second  respectively.  The  corresponding  unit  of  mass,  i.e.,  one 
in  which  a  kilogram  force  would  produce  in  one  second  a  veloc- 
ity of  one  metre-per-second,  is  called  herein  a  geekilogram.* 

233.  Relations  Between  Force  Units  and  Between  Mass  Units. 
— The  dyne  and  the  poundal  might  ideally  at  least  be  "preserved  " 
as  follows:  Place  one  gram  (or  one  pound)  on  one  of  the  small 
carriages  of  fig.  181,  and  then  make  the  large  carriage  move  to 
the  right  with  an  acceleration  of  one  cm.-per-sec.-per-sec.  (or 
one  ft.-per-sec.-per-sec.)  and  note  the  reading  of  the  spring- 
balance.     Assuming  the  mass  of  the  small  carriage  compared  to 

*  This  is  a  new  term,  and  the  student  should  remember  that  fact,  for 
it  is  not  now  and  perhaps  never  will  be  current.  The  unit  is  usually 
called  "engineers'  unit  of  mass,"  an  appellation  which  is  on  a  par  with 
"wood-choppers'  unit  of  volume"  (the  cord).  Other  terms  have  been 
proposed,  but  they  have  never  been  adopted.  The  author  is  confident 
that  names  for  gravitation  units  of  mass  are  convenient  and  that  their  use 
tends  to  clearness. 


§11.]  MOTION  OF  A  PARTICLE.  225 

that  of  its  load,  and  its  frictional  resistance  compared  with  the 
balance  reading  to  be  negligible,  the  force  causing  the  stretch  of 
the  spring  is  one  dyne  (or  one  poundal). 

Instead  of  employing  the  foregoing  method,  we  make  use  of 
measurements  made  on  the  acceleration  due  to  gravity,  thus 
comparing  the  dyne  and  poundal  with  standards  or  units  of 
weight.  A  force  equal  to  the  weight  of  a  gram  (or  a  pound)  act- 
ing on  one  gram  (or  one  pound),  as  in  a  falling  body,  produces 
an  acceleration  of  approximately  981  cm.-per-sec.-per-sec.  (or 
32.2  ft.-per-sec.-per-sec.) ;  hence  a  force  equal  to  1/981  of  the 
weight  of  a  gram  (or  1/32.2  of  the  weight  of  a  pound)  would 
produce  in  one  gram  (or  one  pound)  an  acceleration  of  one  cm.- 
per-sec.-per-sec.  (or  one  ft.-per-sec.-per-sec).  Now  by  defini- 
tion the  forces  producing  in  one  gram  and  in  one  pound  acceler- 
ations of  one  cm.-per-sec.-per-sec.  and  one  ft.-per-sec.-per-sec. 
are  the  dyne  and  poundal  respectively;   therefore 

.    one  dyne  =  1/981  ±  gram  (force), 
one  poundal  =  1/32.2  ±  pound  (force).* 

The  "geepound''  and  the  "geekilogram''  might  also  be  deter* 
mined  by  means  of  the  apparatus  represented  in  fig.  181.  Thus 
adjust  a  load  on  the  smaller  carriage  so  that  the  spring-balance 
will  read  one  pound  (or  one  kilogram)  when  the  larger  carriage 
is  drawn  to  the  right  with  an  acceleration  of  one  ft.-per-sec.-per- 
sec.  (or  one  m.-per-sec.-per-sec).  If  the  mass  and  frictional 
resistance  of  the  small  carriage  are  negligible,  the  mass  of  the 
load  is  one  geepound  (or  one  geekilogram). 

A  better  determination  of  these  units  of  mass  can  be  made 
by  means  of  experiments  on  the  acceleration  due  to  gravity.  A 
one-pound  (or  a  one-kilogram)  force  produces  in  a  one-pound 
(or  a  one-kilogram)  mass  an  acceleration  of  approximately  32.2 
ft.-per-sec.-per-sec.  (or  9.81  m.-per-sec.-per-sec);  hence  a  one- 
pound  (or  a  one-kilogram)  force  would  produce  in  32.2  pounds 
(or  9.81  kilograms)  mass  an  acceleration  of  i  ft.-per-sec  per-sec 
(or  one  m.-per-sec.-per-sec).  Now  by  definition,  two  masses 
which  under  the  action  of  forces  of  one  pound  and  one  kilogram 

*  One  gram  and  one  pound  (force)  being  the  weight  of  one  gram  and 
one  pound  (mass)  respectively. 


2  26  -         MOTION  OF  A  P ARTICLE.  [Chap.  X. 

receive  accelerations  of  one  ft.-per-sec.-per-sec.  and  one  m.-per- 
sec.-per-sec.  are  tl;ie  geepound  and  Idpie  geekilogram  respectively. 
Therefore 

I  geepound  =  3 2. 2  ±  pounds, 

I  geekilogram  =  9.81  ±  kilograms. 

234.  Relation  Between  the  Mass  and  the  Weight  of  a  Body. — 

This  relation  is  implicitly  given  in  the  preceding  article;  we  will 
now  state  it  definitely.  Let  W  denote  the  weight  of  a  body, 
m  its  mass,  and  g  its  acceleration  due  to  gravity,  ike  three  quan- 
tities being  expressed  in  units  of  any  one  kinetic  system.  When 
this  body  falls,  its  acceleration  {g)  is  due  to  its  weight  iW); 
hence  the  equation  of  the  motion  is  (see  art.  231) 

W  =  mg,     or     m  =  W/g, 

and  this  gives  the  relation  between  the  weight  and  the  mass  of 
any  body  when  they  {W,  m,  and  g)  are  expressed  in  units  of  any 
one  kinetic  system.     Thus  for  any  body 

W  (in  dynes)  =  w  (in  grams)  X981  ±  , 

W  (in  poundals)  =  m  (in  pounds)  X  3  2 . 2  ± , 
m  (in  geepounds)  =  W  (in  pounds)  ^  3  2 . 2  i , 
m  (in  geekilograms)  =  W  (in  kilograms)  -^  9.81  ± 

The  weight  and  mass  of  a  body  are  also  numerically  the . 
same  if  expressed  in  certain  units,  more  definitely  if  the  unit 
weight  is  the  weight  of  the  unit  mass  (see  art.  15).     Thus  the 
weight  of  a  barrel  of  flour  is  196  pounds  and  its  mass  is  also  196 
pounds. 

EXAMPLE. 

Express  the  mass  of  a  cubic  foot  of  water  in  pounds,  gee- 
pounds,  kilograms,  and  geekilograrns. 

235.  Acceleration  of  a  Particle  Acted  Upon  by  Several  Forces. 
— The  acceleration  of  a  particle  acted  upon  by  several  forces 
may  be  determined  in  several  ways : 

(a)  By  the  methods  of  '  *  Statics  ' '  determine  the  resultant  of 
the  forces  and  then  compute  the  acceleration  due  to  this  resultant. 
This  acceleration  is  identical  with  that  due  to  the  actual  forces 
applied  to  the  particle,  for  by  definition  the  resultant  of  any 
number  of  forces  is  equivalent  to  them  in  producing  motion. 


§11.]  MOTION  OF  A  PARTICLE.  227 

(6)  Compute  the  acceleration  of  the  particle  due  to  each 
force  acting  alone  and  add  these  accelerations  vectorially.  The 
vector  sum  'represents  the  actual  acceleration  due  to  the  com- 
"bined  action  of  the  forces.  The  correctness  of  this  method  may 
be  proved  from  the  first;  we  give  the  proof  for  the  case  of  two 
fcices:  Let  m  be  the  mass  of  the  particle,  Fj  an^.  Fj  the  applied 
forces,  a  the  acceleration  due  to  their  combined  action,  R  their 
resultant,  and  a^  and  a^  the  accelerations  due  to  F^  and  F^  acting 
singly.     Let  AO  and  BO  (fig.  182)  represent  Fj  and  Fj  respect- 


ively; then  the  diagonal  of  the  parallelogram  OABC  repre- 
sents R  (art.  20),  and  the  continuation  Oc  of  that  diagonal 
represents  a  if  Oc_^R/m  (art.  231).  Also,  Oa  and  Ob  represent 
a^  and  a2  respectively  if 

Oa  =  FJm     and     Ob  =  FJm. 

It  remains  to  show  that  Oc  is  the  vector  sum  of  Oa  and  Ob.  To 
do  this,  join  a  and  c  and  b  and  c,  and  show  that  Oabc  is  a  par- 
allelogram. 

236.  Equations  of  Motion  of  a  Particle. — Let  m,  a,  and  R  de- 
note the  mass  and  acceleration  of  a  particle  and  the  resultant  of 
the  forces  applied  to  it  respectively.  Then,  as  explained  in  the 
preceding  article, 

R  =  ma,   .     .    *. (i) 

and  a  and  R  have  the  same  direction. 

Let  F',  F" ,  etc.,  denote  the  forces  applied  to  the  particle  and 
a,  /?,  and  y  the  angles  which  R  and  a  make  with  a  set  of  coordi- 
nate axes  X,  y,  and  z.  Also  let  ^F^,  IFy,  and  IF^  denote  the 
algebraic  sums  of  the  x,  y,  and  z  components  of  F',  F",  etc.,  and 
ax,  ay,  and  a^  the  x,  y,  and  z  components  of  a.     Now 


228                     MOTION  OF  A  SYSTEM  OF  PARTICLES.         [Chap.  X.  \ 

\ 
R  COS  a=ma  cos  a,    Rcos  ^  =  ma  cos  ^,   and   i^  cos  ;- =  wa cos  ^ ; 

\ 

and  since  -R  cos  a  =  IF^,     R  cos  ^  =  ^Fy,    Rcos  j'  =  IFg, 

and             acosa  =  aa;,          acos/?  =  ay,         acos;'  =  a2,  { 
IF^^msi^,     IFy  =  ma,y,     IF^=msi^.     ...     (2) 

Let  (J)  denote  the  angle  between  the  action  line  of  R  and  the  ■ 

tangent  to  the  path  of  the  particle.     Then  j 

R  cos  (l)  =  ma  cos  (j)     and     R  sin  (f)  =  ma  sin  ^,  ; 

or  IFi  =  meLi    and     i'F„=man (3) 

JFt  and  IFn  denoting  the  algebraic  sums  of  the  tangential  and  1 

normal  components  of  the  forces,  and  at  and  an  the  tangential  .j 

and  normal  acceleration  of  the  particle. 

1 

§  III.     Motion  of  a  System  of  Particles.  l 

237.  Definitions. — Any  number  of  particles  collectively  con- 
sidered are  called  a  system  of  particles.     If  the  distances  between 

the  particles  remain  invariable,  the  system  is  called  a  rigid  one  \ 
or  a  rigid  body. 

Among  the  forces  exerted  upon  the  system  of  particles,  some 

may  be  exerted  by  particles  not  belonging  to  the  system.     Such  ] 

forces  have  already  been  named  external  forces  (art.  113),  and  all  - 

such  forces  the  external  system.      A  force  exerted  by  a  particle  \ 
upon  another  of  the  same  system  has  been  named  an  internal 

force  (art.  113),  and  all  such  forces  the  internal  system.  \ 

According  to  the  third  law  of  motion  (art.  230) ,  if  one  particle 

of  a  system  exerts  a  force  upon  another,  the  second  also  exerts  ; 

one  upon  the  first,  and  these  two  forces  are  equal,  opposite,  and  i 

collinear.     Hence  the  internal  forces  of  a  system  of  particles  \ 

occur  in  pairs,  the  forces  of  each  pair  being  equal,  opposite,  and  ; 
collinear. 

By  effective    force    for    a  particle    of    a  system   is    meant  i 

the  resultant   of  all  the  forces   acting  on  that  particle.      The  i 

effective   forces  for  all    the   particles    of   a   system  are  called  \ 

the  effective  system.      If   m   and  a  denote   the    mass  and  ac-  , 


§111].  MOTION  OF  A  SYSTEM  OF  PARTICLES,  229 

celeration  of  any  particle  of  a  system,  then  (art.  236)  for  that 
particle  the 

effective  force  =  ma. 

238.  D*Alembert*s  Principle. — Since  the  effective  system  con- 
sists of  the  external  and  the  internal  forces  acting  upon  the  par- 
ticles of  a  system,  the  resultant  of  the  effective  forces  is  identical 
with  that  of  the  external  and  the  internal  forces.  Now  the 
resultant  of  the  internal  forces  is  zero  since  they  occur  in  pairs, 
the  forces  of  each  being  equal,  opposite,  and  collinear;   hence 

Proposition  L — For  any  system  of  particles  the  resultants 
of  the  effective  and  the  external  forces  are  identical.  Obviously 
this  proposition  is  equivalent  to 

Proposition  II. — For  any  system  of  particles  the  effective 
forces  reversed  *  and  the  external  forces  together  are  in  equilib- 
rium. This  proposition  is  known  as  D'Alembert's  Principle. 
after  him  who  first  announced  it  (1742)  for  rigid  bodies. 

These  propositions  are  not  fundamental,  being  deducible 
from  the  laws  of  motion  (art.  230),  but  they  express  an  impor- 
tant relation  in  convenient  forms. 

239.  Component  of  an  Effective  System  Along  Any  Line. — 
Let  the  line  be  taken  as  an  x  axis,  and  as  before  let  m^,  m^,  etc., 
denote  the  masses  of  the  particles  and  a/,  a/',  etc.,  the  x  com- 
ponents of  their  accelerations.  Then  the  algebraic  sum  of  the 
X  components  of  the  effective  forces  for  the  particles  is 

Wia^'  +  mjax"  + .  .  . 

This  sum  equals  the  product  of  the  mass  of  the  system  (Im)  and 
the  X  component  of  the  acceleration  of  its  mass-centre  (a^). 
For,  according  to  art.  228, 

m^x^-\-m2X2+  .  .  .  =Jm'X, (i) 

hence  m^dxJdt-^-m^dxJdt^-  .  .  .  =Im'dx/dt,     .     .     .  (2) 

and  m^d'^xJdt'^-\-m2(PxJdt'^-\-  .  .  .  =  Im-d'^x/dt'^^  .     .  (3) 

or  m^aj ^m^ax" \  ...  ^Im-a^ (4) 

240.  Motion  of  the  Mass-Centre  of  any  System  of  Particles. — 

According  to  Prop.  I,  art.  238,  the  algebraic  sums  of  the  com- 

*  Often  called  the  "reversed  effective  system." 


230  MOTION  OF  A   SYSTEM  OF  PARTICLES,  [Chap.  X 

ponents  of  the  external  and  effective  forces  for  any  system  of 
particles  along  any  line  are  equal.  Jtlence,  if  IFx,  ^Fy,  and 
IFz  denote  the  sums  of  the  components  of  the  external  forces 
along  an  x,  y,  and  z  axis  respectively, 

i'F<,  =  2'm-aa:,     lYy^Imsiy,     i'F^  =  i'm•a^.     .     (i) 

These  equations  show  that  the  acceleration  of  the  mass-centre 
depends  only  on  the  values  of  the  components  of  the  external 
forces  and  that 

The  acceleration  of  the  mass-centre  of  a  system  is 
just  like  that  of  a  particle,  whose  mass  equals 
that  of  the  system,  acted  on  by  forces  equal  and 
parallel  to  the  external  forces  which  are  applied 
to  the  system. 

EXAMPLES. 

I.  Show  that  if  the  air  resistance  were  zero  the  acceleration 
of  the  mass-centre  of  any  body  thrown  into  the  air  in  any  way 
would  equal  g  and  be  directed  vertically  downward. 

Solution:  Let  W  denote  the  weight  of  the  body,  and  refer 
the  motion  to  a  set  of  axes  fixed  in  the  earth,  the  y  axis  being 
vertical,  its  positive  end  being  up.  Then  since  the  only  force 
acting  on  the  body  during  the  motion  is  its  weight, 

2F,^o  =  {W/g)a,,  IFy=-W=^{W/g)ay,  IF,  =  o  =  iW/g)a,, 
or  ax  =  ag  =  o,     and     ay=— g;   hence  a=—^. 

*2.  The  mass-centre  of  a  "stationary"  steam-engine  when 
running  is  in  general  not  a  fixed  point.  Show  that  the  reac- 
tion of  its  supports  is  not  constant  in  amount. 

Solution :  Let  m  and  W  denote  the  mass  and  weight  respect- 
ively of  the  engine,  and  a  the  acceleration  of  its  mass-centre. 
Also  let  Rxy  Ry,  and  Rz  denote  the  x,  y,  and  z  components  of 
the  reaction  of  the  supports,  the  y  axis  being  taken  vertical. 
Then  eqs.  (i)  become 

Rx^niGx,    Ry=+W  -{-mdy,    Rz  =  mag,  '^ 

3.  What  is  the  greatest  acceleration  which  a  locomotive  can 
give  a  train  ? 


^  III.]  MOTION  OF  A  SYSTEM  OF  PARTICLES.  231 

Solution:  When  the  locomotive  is  pulling,  the  drivers  tend 
to  slip  on  the  rails  and  the  latter  therefore  exert  on  the  former 
frictional  forces  directed  forward.  (These  may  be  regarded  as 
the  forces  which  directly  make  the  train  move.)  The  rails  exert 
on  the  other  wheels  of  the  train  forces  directed  backward ;  call 
the  sum  of  their  horizontal  components  R\  and  the  resistance 
of  the  air  R''.  If  F  denotes  the  sum  of  the  frictional  forces,  m 
the  mass  of  the  train,  and  a  its  acceleration, 

F-R'-R"  =  ma,     or     a  =  {F-R' -R")/m. 

F  is  maximum  when  the  drivers  are  about  to  slip,  R"  de- 
pends on  the  velocity,  and  is  least  when  the  velocity  is  zero, 
and  R^  does  probably  not  depend  much  on  the  velocity  and 
is  practically  independent  of  the  acceleration.  Hence  the 
acceleration  is  greatest  at  low  speeds  if  the  drivers  are  about  to 
slip. 

4.  What  can  you  say  of  the  forces  which  a  travelling  crane, 
moving  with  an  acceleration,  exerts  on  its  track? 

241.  Moment  of  the  Effective  System  About  Any  Axis. — Let 
the  moment  axis  be  taken  as  an  x  coordinate  axis,  and  call  the 
mass  of  any  particle  m,  its  acceleration  a,  its  coordinates  x,  y, 
and  z.  The  effective  force  for  the  particle  is  ma,  and  its  x,  y, 
and  z  components  are  max,  may,  and  maz  respectively,  and  the 
moment  of  the  force  about  the  x  axis  equals  the  sum  of  the 
moments  of  its  components  (Prop.  I,  art.  28),  i.e., 

ma^ '  y  —  may  •  z. 

The  moment  of  the  entire  system  about  the  x  axis  equals  the 
sum  of  all  such  expressions  as  the  above,  or 

I{mazy  —  mayZ). 

242.  "  Angular  Motion"  of  a  System  of  Particles. — According 
to  Prop.  I,  art.  238,  the  moment  sums  of  the  external  and  effect- 
ive force  systems  about  any  axis  are  equal.  Hence  if  IMx, 
I  My,  and  IMg  denote  the  moment  sums  of  the  external  forces 
about  the  x,  y,  and  z  coordinate  axes  respectively, 


232  MOTION  OF  A  SYSTEM  OF  PARTICLES.  [Chap.  X. 

JM^  =  2(ma^y  -  ma^^z) , 

lUy  =  i'Cma^z  - ma,x) ,  ^  ^,^ 

2'Mg  =  2'  (ma^x — ma-cy ) . 

It  can  be  shown  that  these  and  eqs.  (i)  art.  240  completely 
determine  the  effect  of  a  force  acting  upon  a  rigid  body.  Since 
the  components  and  the  moments  of  a  force  do  not  depend  on  its 
application  point,  the  equations  also  do  not ;  hence  the  effect  of  a 
force  on  the  motion  of  a  rigid  body  does  not  depend  on  its  appli- 
cation point.     This  is  called  the  "  principle  of  transmissibility." 


CHAPTER  XI. 
TRANSLATION  OF  A  RIGID  BODY  (RESUMED). 

§  I.     General  Principles. 

243.  Equations  of  Motion. — Let  m  denote  the  mass  of  a 
body,  a  its  acceleration,  and  ax,  ay,  and  a^  the  x,  y,  and  z  com- 
ponents of  a.  Then,  from  the  equations  of  motion  of  the  mass- 
centre  of  any  system  having  any  motion  (art.  240), 

IFx=m8Lxy     IFy^msLy,     IFg=mgLg.     .     .     .     (i) 

^Fx,  2Fyj  and  IFg  denoting  the  algebraic  sums  of  the  x,  y, 
and  z  components  of  the  external  forces.  It  is  advantageous 
to  select  the  coordinate  axes  in  a  certain  way  in  special  cases ; 
thus  if  the  motion  is  a  plane  one,  two  axes  should  be  taken  in 
the  plane  of  the  motion,  for  then  the  equations  reduce  to  two  in 
number;  and  if  the  motion  is  rectilinear,  one  of  the  axes  should 
be  taken  parallel  to  the  direction  of  the  motion,  for  then  the 
number  of  equations  reduces  to  one. 

It  is  shown  in  the  next  article  that  the  resultant  of  the  effect- 
ive forces  for  the  particles  of  a  translating  body  is  a  single  force 
acting  through  the  mass-centre  in  the  direction  of  the  accelera- 
tion, its  magnitude'  being  equal  to  the  product  of  the  mass  of 
the  body  and  its  acceleration.  According  to  D'Alembert's 
Principle,  the  resultants  of  the  external  and  effective  systems 
are  identical;  hence  if  the  resultant  of  the  external  forces  is 
denoted  by  R, 

R=ma (2) 

and  it  acts  through  the  mass-centre  in  the  direction  of  the  accel- 
eration. 

244.  Resultant  of  the  Effective  System. — In  a  translation, 
the  accelerations  of  all  the  particles  of  the  moving  body  at  each 
instant  are  alike  in  magnitude  and  in  direction.  Hence  the 
effective  system  consists  of  forces  having  the  same  direction, 

233 


^34  TJOINSLATION  OF  A  RIGID  BODY,  [Ch.  XL 

and  they  are  proportional  to  the  masses  of  the  corresponding 
particles;  therefore  the  resultant  of  the  effective  system  is  a 
single  force. 

Obviously  the  direction  of  the  resultant  at  each  instant  is 
the  same  as  that  of  the  acceleration.  The  magnitude  equals  the 
sum  of  the  separate  effective  forces,  i.e., 

{dm)^a^  +  {dm)2a2  +  .  .  .  =  a  J  dm  =  ma, 

(c?m)i,  (dm)^,  etc.,  denoting  the  masses  of  the  particles.  The 
action  line  of  the  resultant  passes  through  the  mass-centre,  as 
can  be  shown  thus:  The  effective  forces  constitute  a  system  of 
parallel  forces  with  fixed  application  points ;  hence  they  have  a 
centre  or  centroid  (art.  62).  Let  x,  y,  and  z  denote  the  coordi- 
nates of  any  particle  with  reference  to  a  set  of  fixed  axes,  x^,  y^, 
and  Zq  the  coordinates  of  the  centroid  of  the  effective  forces,  and 
Xy  y,  and  Z  the  coordinates  of  the  mass-centre.     Then  (see  art.  63) 


/  (dm '  a)x      I  dm  -  x 


Xq= = =  x     (see  art.  228). 

^  ma  m  ' 

Similarly  it  can  be  shown  that  y^^^y  and  Zq  =  z\  hence  the  cen- 
troid of  the  effective  forces  coincides  with  the  mass-centre,  and 
the  resultant  of  the  effective  forces  passes  through  the  mass- 
centre  as  stated. 

§  II.     Applications. 

245.  General  Method  of  Procedure. — In  the  following  prob- 
lems the  forces  applied  to  a  body  are  wholly  or  partially  given 
and  it  is  required  to  determine  the  acceleration,  or  else  the 
acceleration  is  given  and  one  or  more  forces  are  required.  Such 
problems  are  solved  by  writing  the  equations  of  motion  (or  as 
many  as  necessary)  and  then  solving  them  for  the  unknowns. 

If  it  is  required  to  determine  the  motion  completely,  i.e.,  to 
compute  the  acceleration,  velocity,  and  position  at  each  instant 
of  the  motion,  the  acceleration  is  found  first  as  just  indicated 
and  then  the  velocity  and  position  may  be  found  by  methods 
explained  in  Chapters  VII  and  VIII. 


§II.j  APPLICATIONS.  235 

EXAMPLES. 

I.  If  the  "body  represented  in  fig.  183(a)  weighs  50  lbs.,  the 
f)uii  P  is  40  lbs.,  the  angle  ^  is  o,  and  the  supporting  surface  is 

:.p  lbs. 

40  lbs. 


"PR 

(6) 


Fig.  183. 


smooth,  compute  the  acceleration  and  the  reaction  of  the  sup- 
port. 

Solution :  Let  R  denote  the  reaction  of  the  plane ;  then  the 
external  forces  acting  on  the  body  are  as  represented  in  fig. 
183(6).  The  mass  of  the  body  is  50/32.2  =  1.553  geepounds. 
Hence 

i'F^  =  4o  =  i.553aa;, 

IFy^R-So  =  i.ssiay, 

2*^^  =  0  =  I  ;553a^. 

From  the  third  equation  0^  =  0,  and  from  the  first  a^^  25.76 
ft. /sec. ^  Obviously  ay  =  o,  hence  the  acceleration  of  the  body 
is  25.76  ft. /sec. ^  in  the  plus  x  direction.  Since  ay  equals  zero, 
the  second  equation  shows  that  R  =  So  lbs. 

^  2.  Suppose  that  at  a  certain  instant  the  mass-centre  of  the 
body  of  ex.  i  is  at  the  origin,  its  velocity  being  30  ft.-per-sec.  in 
the  plus  X  direction.  Compute  the  velocity  and  position  of  the 
mass-centre  two  seconds  later. 

,  3.  Suppose  that  at  a  certain  instant  the  mass-centre  of  the 
body  of  ex.  i  is  at  the  origin  and  that  the  velocity  of  the  body 
at  that  instant  is  30  ft.-per-sec.  in  the  plus  z  direction.  Deter- 
mine the  velocity  and  the  position  of  the  mass-centre  two  seconds 
later. 

^    4.  Solve  ex.  i,  supposing  that  6  equals  20°. 

Ans.  a  =  24.2  ft. /sec/sec. 
5.  Suppose  that  P  =  40  lbs.  and  ^  =  0  (fig.  183),  that  the  sup- 
porting surface  is  rough,  the  frictional  resistance  being  10  lbs., 


236 


TRANSLATION  OF  A  RIGID  BODY. 


[Ch.  XL 


and  that  at  a  certain  instant  the  body  is  at  rest.     Determine  the  \ 

subsequent  motion.                                  ^.  i 

6.  Suppose  that  at  a  certain  instant  the  body  of  ex.  5  is  \ 
moving  in  the  minus  x  direction  with  a  velocity  of  30  ft. -per-  \ 
sec.     Determine  the  subsequent  motion.              \                           -'^^ 

7.  Suppose  that  the  inclined  plane  (fig.  184a)  is  smooth,  that  j 
the  body  upon  it  weighs  50  lbs.,  and  that  : 
P  equals  40  lbs.     Determine  the  acceler- 
ation and  the  reaction  of  the  plane.  i 

(^)        Solution:    Let  i?  denote  the  reaction  ' 
of  the  plane;    then  the  external  forces 

acting  on  the  body  are  as  shown  in  Hg.  \ 

184(6).     Taking  coordinate  axes  as  shown,  j 

the  mass  being  1.553  geepounds,  ^ 

i'i^a:= -40 +  50  cos  60°  =1.5 53a;c,  : 

2'Fy=i?-5osin6o°  =  i.553aj„  i 

i'F^  =  o  =  i.553a;5.  ] 

From  the  first  and  third  respectively. 


Fig.  184. 


—  9 .66 ,     ag  =  o,     and  obviously     a^  =  o ; 


hence  0  =  9.66  ft. /sec. ^  in  the  negative  x  direction. 

From  the  second  equation  it  follows  that  i^  =  43.3  ^bs. 
.     8.  Suppose  that  the  plane  in  fig.  184(a)  is  rough  and  that  the 
frictional  resistance  is  10  lbs.     If  at  a  given  instant  the  velocity 
of  the  body  is  zero,  determine  the  Subsequent  motion.  \    --■-  • 

9.  Suppose  that  at  the  instant  mentioned  in  the  preceding 
example  the  velocity  is  40  ft.-per-sec.  down  the  plane.  Deter- 
mine the  subsequent  motion. 

10.  Fig.  185(a)  represents  an  open  box  on  a  horizontal  sur- 
face, the  box  containing  a  stone  and  subjected  to  a  force  P.  Let 
the  weights  of  the  box  and  stone  be  90  and  70  lbs.  respectively, 
P  100  lbs.,  and  suppose  that  the  supporting  surfaces  are  smooth. 
Compute  the  pressure  between  the  stone  and  the  rear  end  of  the 
box. 

Solution:  The  external  forces  applied  to  box  and  stone 
together  consist  of  the  pull,  the  reaction  of  the  horizontal  sup- 
port and  the  weights  (see  fig.  1856).     The  masses  of  the  two 


§11.] 


APPLICATIONS. 


237 


bodies  are  70/32.2  and  90/32.2,  or  2.17  and  2.80  geepounds  re- 
spectively. Evidently  the  acceleration  a  is  in  the  direction  of 
the  loo-lb.  force  and  its  value  is  given  by 


100  =  4. 97a,     or 


20.1  ft. /sec. /sec. 


The  external  forces  acting  on  the  box  and  stone  respectively  are 
shown  in  (c)  and  {d) ,  P'  and  R^  denoting  pressures  between  the 
stone  and  the  end  and  bottom  of  the  box  respectively.     Know- 


(a) 


H~' 


Tr 


(0) 


100  lbs. 


h£(e). 


Fig.  185. 

ing  the  acceleration  of  each  body,  the  forces  can  be  determined 
from  the  equations  of  motion  of  either;  thus  for  the  stone, 

P'  =  2.i7  X20.1  =43.62  lbs. 

t^  II .  Four  bodies  are  connected  by  cords  as  shown  in  fig.  i85(^) 
and  are  pulled  along  on  a  horizontal  plane  by  a  force  P  of  200 
lbs.  Supposing  the  frictional  resistance  on  each  body  to  equal 
one-fourth  its  weight,  compute  the  tension  in  each  cord. 
I  12.  In  an  elevator  there  are  two  boxes  weighing  610  and  1000 
lbs.  respectively,  the  lighter  being  on  top  of  the  other.  What 
are  the  pressures  on  the  bottoms  of  the  boxes  if  (a)  the  elevator 
is  started  up  with  an  acceleration  of  4  ft. /sec. /sec?  (b)  If 
started  down  with  the  same  acceleration? 

Ans.  (a)  18 10  lbs.  on  the  bottom  of  the  lower  box. 

13.  If  the  elevator  of  the  preceding  example  weighs  1600 
lbs. ,  compute  the  tensions  in  the  hoisting  cable  in  the  two  cases 
mentioned,    -ai  ?60f.6^     i^j     X'^'c^ 

14.  Fig.  186(a)  represents  two  bodies,  A  and  B,  suspended 
from  the  ends  of  a  cord  which  passes  over  a  "smooth  pulley"  6'. 
Let  the  weights  of  A  and  B  be  W^  and  W^  iyV^  being  the  greater). 


238 


TRANSLATION  OF  A  RIGID  BODY. 


[Chap.  XI. 


Assume  that  the  cord  and  pulley  are  practically  without 
mass  and  that  the  axle  friction  is  zerp;  then  the  tension  is  the 
same  at  all  sections  of  the  cord.  Compute  the  acceleration  of 
the  bodies  and  the  tension. 


^^f^ 


(h) 


<b 


Fig.  i86. 


Solution :  Let  T  denote  the  tension ;  then  the  external  forces 
applied  to  each  body  are  as  represented.  Assuming  the  string 
to  be  inextensible,  the  accelerations  of  the  two  bodies  are  equal; 
we  will  denote  them  by  a.  Evidently  the  accelerations  of  A 
and  B  are  respectively  upward  and  downward;  hence  the  re- 
sultant forces  on  A  and  B  act  up  and  down  respectively,  i.e., 
W^KTkW^.  The  masses  of  A  and  B  being  WJg  and  WJg 
respectively,  the  equations  of  motion  for  these  bodies  are 


g 


and     W,-T  =  ^^a 
g 


Solving  these  equations  for  T  and  a,  we  find  that 


a  = 


W,-W, 


and     T  = 


2W,W, 


(/ 


15.  Fig.  186(6)  represents  two  bodies,  A  and  B,  connected 


by  a  cord  passmg  over  a  smooth  pulley,  one  hanging  freely  and 
the  other  supported  by  a  horizontal  surface.  Let  the  weights 
of  A  and  B  be  Wj^  and  W^  respectively,  and  assume  the  surface 
to  be  smooth  and  that  the  tensions  at  different  sections  of  the 
cord  are  equal.  Compute  the  acceleration  of  the  bodies  and  the 
tension.  • 


/ins.  a  = 


g.     T  = 


I  16.  Solve  the  preceding  ex.,  supposing  that  the  T^i-=64  lbs., 
1^2  =  96  lbs.,  and  that  the  horizontal  surface  is  so  rough  that  its 
frictional  resistance  on  the  body  is  19.2  lbs.     (Take  g  =  32.) 


7 '  ff<ipr  ^  ^ .  r.fS^^^' 


§  II.]  APPLICATIONS.  239 

?  17.  Suppose  that  a  stone  weighing  100  lbs.  is  placed  on  a 
rough  horizontal  board,  the  coefficient  of  static  friction  for  the 
stone  and  board  being  1/4.  If  the  board  is  moved  horizontally 
with  an  acceleration  of  4  ft. /sec. /sec,  will  the  stone  remain  on 
the  board? 

'^  18.  Suppose  that  the  stone  of  the  preceding  example  is  square 
in  cross-section,  i  by  i  ft.,  and  8  ft.  high,  and  that  it  stands  so 
that  two  of  its  sides  are  parallel  to  the  direction  of  the  motion. 
How  large  an  acceleration  of  the  board  will  cause  the  stone  to 
tip,  supposing  that  the  friction  between  the  board  and  stone  to 
be  large  enough  to  prevent  slipping. 

246.  Kinetic  Reactions. —  When  a  body  rests  upon  a  hori- 
zontal support,  the  reaction  between  the  support  and  body 
equals  the  weight,  but  if  the  body  is  in  motion,  the  reaction  in 
general  does  not  equal  the  weight.  Thus,  as  seen  in  ex.  12,  art. 
245,  when  an  elevator  moves  with  an  acceleration  a,  the  pres- 
sure which  it  exerts  upon  a  body  and  its  floor  equals  W ±ma 
(W,  m,  and  a  denoting  the  weight,  mass,  and  acceleration  of  the 
body  respectively),  the  plus  or  minus  sign  being  used  according 
as  the  acceleration  is  up  or  down.  The  reactions  of  the  cranks 
on  a  coupling-rod  of  a  locomotive  when  it  is  at  rest  each  equal 
one-half  the  weight  of  the  rod,  but  when  it  is  in  motion  they 
have  a  different  value.  The  difference  depends  in  part  on  the 
acceleration  as  is  shown  below. 

The  components  of  a  force  acting  on  a  body  which  depend 
upon  acceleration  are  said  to  be  "due  to  acceleration"  and 
**due  to  inertia,"  and  such  are  often  called  inertia  forces.  Some- 
times it  is  desirable  to  determine  the  components  of  a  reaction 
which  are  dependent  and  independent  of  acceleration.  To  dis- 
tinguish them  we  shall  call  the  former  and  latter  kinetic  and 
static  reactions  respectively.  No  new  principles  are  necessary 
for  the  computation  of  these. 


EXAMPLES. 

I.  Suppose  that  the  mass  of  the  slider  represented  in  fig.  187 
is  m,  the  length  of  the  crank  c,  and  the  number  of  revolutions 
per  unit  time  n.     Compute  the  crank-pin  pressure  at  several 


240 


TRANSLATION  OF  A  RIGID  BODY. 


[Chap.  XL 


points  of  the  stroke  and  plot  curves  showing  how  it  varies  with 
the  displacement  and  the  time.  »^ 

Solution :  Neglecting  friction  the  only  horizontal  force  on  the 
slider  is  the  crank-pin  pressure.  Hence,  denoting  that  pressure 
by_Q  and  the  acceleration  of  the  slider  bv  a, 

the    positive    direction  being  taken   the   same   for  Q  and  a. 

Since  a=  —^itVc  sin  d=  —47tVx  (see  eq.  3,  art.  179), 

Q=  —nt^nVc  sin  d=  —m47tVx; 

and  the  maximum  value  of  Q  is  given  by 
Q^=-ni47tVc. 

Fig.  187  (6)  and  (c)  shows  graphically  how  Q  varies  with  the  dis- 
placement and  the  time. 


/ 

/ 

1    ^  ^"' 

.    • 

V 

\ 

1 

1 
I 

v.. 

9 
• 

Fig.  187. 

2.  Assume  that  the  slider  of  the  preceding  example  is  verti- 
cal, that  m  =  i8o  lbs.,  c  =  to  in.,  n  =  i20  rev.-per-min.  Draw 
curves  similar  to  those  of  fig.  187,  showing  the  static  and  kinetic 
components  of  the  entire  pin  reaction. 

3.  Assume  that  the  slider  of  fig.  187  is  driven  by  steam  pres- 
sure P,  its  value  being  known  at  each  point  of  the  stroke.  As- 
suming that  the  crank  turns  uniformly,  determine  the  crank- 
pin  pressure  for  any  position. 

Solution:  (i)  Algebraic.     With  notation  as  in  ex.  i, 
P  +  Q=ma     or     Q  =  ma  —  P, 

a  being  equal  to  —47?^^^^  sin  d  or  —47rVx,  as  in  ex.  u 


i    jn.] 


APPLICATIONS. 


241 


(2)  Graphical  and  employing  D'Alembert's  Principle.  Since 
P,  Qy  and  the  reversed  effective  force  for  the  slider  are  at  all 
times  in  equilibrium  (art.  238),  Q  is  equal  and  opposite  to  the 
resultant  of  P  and  the  reversed  effective  force. 

Let  aV  (fig.  188)  represent  the  length  of  the  stroke,  and  the 


Fig.  188. 


ordinates  from  a'a"  to  the  curve  c'c"  the  values  of  the  steam 
pressure,  P.  From  a'  to  a  the  pressure  is  forward  and  from  a 
to  a"  backward.  The  effective  force  for  the  slider  at  each  instant 
equals  the  product  of  its  mass  and  acceleration ;  it  varies  there- 
fore with  the  acceleration,  and  the  ordinates  from  a'a''  to  the 
straight  line  h'h"  may  represent  the  effective  forces.  During 
the  first  part  of  a  stroke  the  acceleration  is  forward  (in  the  direc- 
tion of  the  motion)  and  in  the  second  part  backward ;  hence  the 
reversed  effective  force  in  the  first  part  of  the  stroke  is  backward 
and  in  the  second  forward. 

The  directions  of  the  steam  pressure  and  reversed  effective 
force  from  a'  to  h  are  opposite,  from  6  to  a  the  same,  and  from 
a  to  a"  opposite  again.  Hence  the  resultants  of  the  two  forces 
from  a'  to  h  and  from  a  to  a"  are  represented  by  the  differences 
between  the  ordinates  to  the  two  curves,  and  from  6  to  a  by  their 
sum,  i.e.,  by  the  ordinates  of  the  shaded  area.  Obviously  from 
a'  to  c  the  direction  of  the  horizontal  pin  pressure  is  opposite  to 
the  direction  of  the  motion  and  from  c  to  a"  in  that  direction. 

4.  Compute  the  kinetic  reactions  on  a  coupling  or  side  rod 
of  a  locomotive  running  at  a  constant  speed. 
Solution :  Let  the  notation  be  as  follows : 
r,  radius  of  the  crank-pin  circles,    v,  speed  of  the  locomotive, 
R,     "       **    '*    drivers,  m,  mass  of  the  rod. 

Since  (by  supposition)  the  acceleration  of  the  locomotive  is  zero, 


242  TRANSLATION  OF  A  RIGID  BODY,  [Chap.  XI. 

the  absolute  acceleration  of  the  rod  is  the  same  as  its  accel- 
eration relative  to  the  locomotive  (s^e  Prop.,  art.  196).  Now 
the  motion  of  each  point  of  the  rod  is  just  like  that  of  the 
centre  of  a  crank-pin,  and  relative  to  the  locomotive  that  motion 
y  is  circular,  the  speed  being  rv/R.^  When  a  point  travels  in  a 
circle  with  a  constant  speed,  its  acceleration  equals  the  speed 
squared  divided  by  the  radius  of  the  circle  and  its  direction  is 
along  the  radius  toward  the  centre  (art.  193).  Hence  the  accel- 
eration of  the  rod  equals  rv^/R^,  and  its  direction  at  each  instant 
is  parallel  to  the  crank. 

The  effective  force  therefore  equals  mrv^/R^,  and  if  the  mass- 
centre  is  at  the  middle  of  the  rod  each  kinetic  reaction  equals 
mrv'^/2R'^,  its  direction  being  from  the  corresponding  crank-pin 
to  the  centre  of  the  wheel. 

The  total  crank-pin  pressure  depends  also  on  the  weight  of 
the  rod  and  on  the  train  resistance  which  is  being  "overcome." 
*^'*  5.  Assume  that  the  weight  of  one  side  rod  is  275  lbs.,  crank 
radius  i  ft.,  wheel  diameter  5^  ft.,  and  the  speed  60  mi.-per-hr. 
Compute  the  crank-pin  reactions  due  to  acceleration  and  weight 
of  the  rod  when  it  is  at  its  lowest,  highest,  and  middle  posi- 
tions. 

247.  Vibrations. — A  study  of  vibrations  furnishes  applica- 
tions of  equations  of  motion  in  which  the  force  is  variable.  As 
illustrations,  we  choose  vibrations  caused  by  open  coil-springs  in 
various  circumstances.  If  such  a  spring  hangs  vertically  from 
one  end  and  supports  a  body  at  the  other,  and  the  body  is  dis- 
placed vertically  from  its  position  of  rest  and  then  released,  it 
will  oscillate  or  vibrate  up  and  down,  the  vibration  enduring  for 
a  time  and  then  ceasing.  This  is  called  a  natural  vibration,  and 
in  general  a  natural  vibration  is  one  executed  by  a  body  or  sys- 
tem which  has  been  displaced  or  distorted,  then  released  and 
left  to  itself. 

If  the  support  of  the  coil-spring  is  not  fixed,  but  has  a  periodic 
up  and  down  motion,  the  motions  of  the  spring  and  suspended 
body  are  called  forced  vibrations.  In  general  if  a  body  or  system 
executes  a  natural  vibration  and  is  then  subjected  to  a  periodic 
influence  on  its  motion,  the  resulting  vibration  is  called  a  forced 
one. 


§IL] 


APPLICATIONS. 


243 


The  dying  out  of  a  vibration  is  ascribed  to  forces  of  the  nature 
of  friction ;  thus  in  the  preceding  illustrations  the  forces  are  the 
resistance  of  the  air  upon  the  moving  bodies  and  the  internal  or 
molecular  friction  in  the  spring.  This  effect  (dying  out)  is 
known  as  damping,  and  vibrations  in  which  damping  occurs  are 
called  damped  vibrations.  For  simplicity  we  will  assume  that 
there  are  springs  without  internal  friction,  and  hence  that  such 
a  spring  and  a  suspended  body  might  execute  an  undamped 
(or  "simple")  vibration. 

If  an  open  coil-spring  is  elongated  or  compressed  (not  exces- 
sively), the  elongating  or  compressing  force  is  proportional  to 
the  elongation  or  compression  of  the  spring.  Strictly,  this  is  true 
only  when  the  act  of  elongation  or  compression  is  slow ;  we  will 
3,ssume  it  to  be  true  for  a  vibrating  spring.  No  important  error 
results  if  the  mass  of  the  suspended  body  is  considerable  as  com^ 
pared  with  that  of  the  spring.  Then  if  e  and  e'  denote  the  elon- 
gations or  compressions  of  a  spring  caused  by  forces  T  and  T' 
respectively, 

T/T=e/e'     or     T  =  {T'le')e. 

248.  Undamped  Natural  Vibration. — We  use  the  following 
notation  (see  fig.  189): 

/  =  natural  length  of  the  spring; 
W^  =  weight  of  the  suspended  body; 

m  =  mass      "     "  "  '* 

^'  =  elongation  caused  by  W\ 

k  =  W/e'\ 

y  =  displacement  of  the  body  from  its  position  of  rest ; 

a  =  acceleration     '*    **       "     at  the  displacement  >/; 

T  =  force  exerted  upon  the  body  by  the  spring  at  the 

displacement  y. 
We  choose  the  downward  direction  as  positive  for 
forces  and  displacements. 

The  elongation  (or  contraction)  of  the  spring  for 
all  positions  of  the  body  is  {y-\-e') ;  the  spring  is  elon- 
gated or  compressed  according  as  {y-\-e')  is  positive 
or  negative.     Also 


Fig.  189. 


T=-iyV/e'){y-\-e')  =  -{W/e)y''W, 


244  TRANSLATION  OF  A  RIGID  BODY.  [Chap.  XI. 

and  the  resultant  force  on  the  body  in  any  position  is 

W+T==-{W/e')yrr-ky (i) 

Hence  the  equation  of  motion  is 

—  ky  =  ma,  or  a=  —{k/m)y,  ....  (2) 
i.e.,  the  acceleration  is  proportional  to  y  (displacement)  and  is 
always  directed  toward  a  fixed  point  (from  which  y  is  measured) 
in  the  path;  therefore  the  motion  is  simply  harmonic  (see  art. 
179)  and  its  period  is  27r/\/k/m. 

EXAMPLE. 

Suppose  that  PF  =  5  lbs. ,  ^'  =  3  in. ,  and  that  the  body  is  released 
from  a  position  10  in.  below  that  of  rest.     Describe  the  motion. 

249.  Damped  Natural  Vibration. — The  laws  of  the  damping 
forces  are  various,  depending  upon  the  circumstances  of  the 
motion.  If  the  vibrating  system  is  like  that  represented  in  fig. 
189,  the  suspended  body  being  a  thin  vertical  plate  immersed  in  a 
viscous  liquid,  then  (neglecting  the  internal  friction  in  the  spring) 
the  damping  force  consists  of  the  fluid  friction  at  the  sides  of 
the  plate.  If  the  velocity  of  the  plate  is  small,  this  friction  is  ap- 
proximately proportional  to  the  velocity.  We  choose  this  law 
of  damping  because  the  corresponding  motion  is  analogous  to  an 
important  electrical  phenomenon. 

Let  V  and  v'  denote  any  two  velocities  of  the  immersed  plate 
and  F  and  F'  the  corresponding  frictions ;  then 

F/F'  =  v/v'     or     F={F'/v')v  =  cv, 

c  being  an  abbreviation  for  F' jv' .  If  F  be  regarded  as  positive 
or  negative  according  as  it  acts  down  or  up  on  the  plate,  then 
because  v  =  dy/dt  (see  the  preceding  article),  F  =  cdyjdt,  and 
the  resulting  force  on  the  plat.e  is 

W  +  T  +  F=-ky-cdy/dt (i) 

The  equation  of  motion  of  the  plate  becomes,  instead  of  eq.  (2) 
art.  248, 

dy         d^y 


—  ky  —  c-fj-  =  m 


dt  dt 


..   ,    -    dy     k 


d^y      c  dy 
dt^     m  dt 


(2) 


§  II.]  APPLICATIONS.  245 

For  brevity,  let  p  =  c/m  and  q  =  Vk/m ;  then  the  solution  of  eq. 
(2)  gives  the  following: 
I  (a)  li\p'<q\ 

y^Ar^'^'sm  {Vq'^-pyA  ^+£),  ....    (3) 

f  £  being  the  "Naperian  base"  and  A  and  s  constants  of  integra- 
tion depending  upon  "initial  conditions."  Thus  let  /  =  o  and 
dy/dt  =  VQ  when  y  =  o,  then  substituting  these  in  eq.  (3)  and  in 
the  expression  for  dy/dt,  we  find  that 

£  =  0     and     A=vJ\/q'^—p'^//^, 

I  {h)li\p'>q\ 

y^Ar'^'+Br^', (4) 

a  and  /?  being  abbreviations  ioT—p/2  ±  V'^V4— g^  respectively 
and  yl  and  B  constants  of  integration  depending  on  initial  con 
ditions.     If  as  under  (a),  t  =  o  and  dy/dt  =  VQ  when  y  =  Oy  then 
substituting  these  in  eq.  (4)  and  in  the  expression  for  dy/dt^ 
we  find  that 

A=v,/{p-a)     and     B==vJ(a-J3), 

EXAMPLE. 

Letg  =  4  rad.-per-sec.  and  z^o^^^  ft.-per-sec.      Plot  on  the 
,     same  axes, 
I  (1)  Equation  (4),  p  being  9  rad.-per-sec. 

(2)  Equation  (3),  ^  being  2  rad.-per-sec. 

(3)  Equation  {3),  p  being  zero. 

P  250.  Undamped  Forced  Vibration. — Imagine  the  support  of 
the  spring  in  fig.  189  to  oscillate  up  and  down  and  call  its  dis- 
placement from  the  position  shown  x,  regarding  x  as  positive  or 
negative,  according  as  the  displacement  is  down  or  up.  The 
elongation  of  the  spring  at  any  instant  is  not  y+e^  as  in  art.  248, 
huty+e^—x;  hence 

I  T=-(W/e^Xy  +  e^-x), 

T    and  the  resultant  force  on  the  suspended  body  is 

W+T=^kiy-x) (i) 


246  TRANSLATION  OF  A  RIGID  BODY,         .     [Chap.  XI. 

The  equation  of  motion  is 

or  (Py/dt''  +  {k/m){y-x)=o] ^^^ 

Now  let  the  motion  of  the  support  be  simply  harmonic,  its 

amplitude,  period,  and  epoch  being  A,  27:/ lo,  and  o  respectively; 

then  eq.  (2)  becomes 

dH 

^  +  q^y  =  q^ A  sin  (ot, (3) 

q  being  an  abbreviation  for  \^k/m,  as  in  art.  249.  The  solution 
of  the  last  equation  is 

y^l-tyq^'"'"'' ■      •      (4) 

This  shows  that  the  forced  vibration  of  the  suspended  body  is 
simply  harmonic,  its  period  and  epoch  being  the  same  as  of  the 
motion  of  the  support,  and  its  amplitude  1/(1— a>V'7^)  times 
that  of  the  support.  Notice  that  co/q  is  the  ratio  of  the  fre- 
quency of  the  motion  of  the  support  to  that  of  the  natural  vibra- 
tion of  the  spring. 

EXAMPLE. 

Let  A  =  i,  and  plot  a  curve  showing  how  the  amplitude  of 
the  forced  vibration  varies  for  different  values  of  cu/q,  between 
i/io  and  10. 

251.  Damped  Forced  Vibration. — Imagine  that  the  suspended 
body  of  the  preceding  article  is  a  vertical  thin  plate  immersed  in 
a  viscous  liquid.  Then  in  addition  to  the  forces  acting  on  the 
body  as  described  in  the  preceding  article,  there  is  the  frictional 
resistance  F-^  —cdy/dt  (see  art.  249).  Hence  the  equation  of 
motion  is 

d^y      c  dy  ^   J^  dy      k    .    . 
dv      m  dt      m  at      m 

d^y       dy       ,         t  a    • 
or  d^^^di^ ^^  ^  ^      ^^^  ^^' 


(I) 


p  and  q,  being  abbreviations  as  in  art.  249.      The  solution  of 
this  equation  is 


y=' A  sva.ia)t  —  e)t (2)     1 


§  II.  ]  APPLICA  TIONS.  247 

£  being  an  abbreviation  defined  by 

tan  £  =  (jup(q^  —  co^). 

This  equation  shows  that  the  motion  of  the  suspended  body  is 
simply  harmonic,  its  period  and  amplitude  being  respectively 
the  same  as  and  {q^  sin  £)/ajp  times  that  of  the  motion  of  the 
support  and  lagging  behind  the  latter  an  amount  equal  to  s. 

EXAMPLE. 

Assume  that  q  =  io  rad.-per-sec.  and  that  (o  varies  from  7 
to  13  rad.-per-sec.  Draw  curves  showing  (a)  how  the  lag  e 
varies  with  (o  for  values  of  p  equal  to  2,  i,  1/2,  and  i/io,  and 
(b)  how  the  amplitude  varies  with  co  for  the  four  values  of  p 
just  given. 

252.  Kinetic  Friction. — Definitions.  The  friction  between 
two  bodies  which  move  relative  to  each  other  is  called  kinetic 
friction,  and  the  ratio  of  the  kinetic  friction  to  the  normal 
pressure  between  the  bodies  is  called  their  coefficient  of  kinetic 
friction.  If  /  denotes  the  coefficient  and  F  and  N  the  friction 
and  normal  pressure  respectively, 

f  =  F/N     or     F  =  /A/' (i) 

Laws  of  Friction  for  Dry  Surfaces. — (i)  The  coefficient  de- 
pends on  the  nature  of  the  rubbing  surfaces. 

(2)  The  coefficient  is  approximately  independent  of  the  in- 
tensity of  the  normal  pressure.  Strictly  it  falls  off  slightly  as 
the  intensity  increases  up  to  a  point  when  "seizing"  is  about  to 
occur;  then  it  increases  rapidly. 

(3)  The  coefficient  decreases  as  the  velocity  increases — not 
directly,  but  rapidly  as  the  velocity  increases  from  o  to  0  + 
and  then  less  rapidly  with  increasing  velocity. 

(4)  The  coefficient  decreases  with  the  lapse  of  time  during 
the  motion. 

These  laws  are  qualitative;  their  reduction  to  a  quantita- 
tive form  has  not  been  effected,  but  in  some  cases  we  know 
quite  accurately  how  the  coefficient  varies,  especially  with 
velocity.     Thus  in  certain  experiments  on  the  braking  of  va^^- 


248  TRANSLATION  OF  A  RIGID  BODY,  [Chap.  XL 

way  trains,  the  following  values  of  the  coefficient  for  the  brake- 
shoe  (cast  iron)  and  the  wheel  (steel>were  determined: 

Velocity  in  mi.-per-hr. .      17        21         27        31        37        47 
Coefficient 0.16     0.15     0.13     o.ii     o.io     0.08 

These  coefficients  were  taken  five  seconds  after  the  brakes  were 
set;  fifteen  seconds  after  setting,  the  coefficients  were  about 
0.04  less  than  those  given  above. 

EXAMPLES. 

^.       I.  Fig.  190(a)  represents  a  body  resting  on  a  plank,  which  in 
turn  rests  on  rollers.     If  f  is  the  coefficient  of  static  friction  for 


u^iumtim»MaJMv/jJi3mug/u/jMXUi^^ 


Fig.  190. 

the  body  and  plank,  how  large  an  acceleration  can  be  given  to 
the  plank  without  causing  slipping  between  it  and  the  body  ? 

Ans.  fg. 
V  2.  Suppose  that  the  body  is  slipping  on  the  plank,  and  let 
f  denote  the  coefficient  of  kinetic  friction  assuming  it  to  be 
independent  of  velocity.  How  long  {t)  and  how  far  {s)  will  the 
body  slide  in  coming  to  rest  from  a  velocity  v  under  the  influence 
of  friction?  Ans.  t  =  vlf'g  and  s  =  v^/2f'g. 

3 .  Suppose  that  the  plank  is  caused  to  oscillate  on  the  rollers 
by  a  force  P,  so  that  the  acceleration-time  curve  for  the  motion 
is  represented  by  fig.  190(6).  Draw  the  velocity-time  and  space- 
time  curves  for  the  motion  of  the  plank;  also  for  the  body,  tak- 
ing f  =  0.5,  j"  =  0.2,  and  g  =  32  ft. /sec. ^  Where  on  the  plank  is 
the  body  at  the  end  of  one  second? 

Ans.  It  has  slid  forward  on  the  plank  25?^  ft. 


CHAPTER  XII. 

ROTATION  (RESUMED). 

i  I.     Second  Moments  of  Mass  (Moment  op  Inertia,  btc). 

253.  Occurrence  of  Second  Moments. — In  a  study  of  the  rota- 
tion of  a  body,  certain  quantities  are  met  with  which  are  ex- 
pressed by  integrals  of  the  kind  and  form  /  dm-u^  and  /  dm-uv, 

m  denoting  mass  and  u  and  v  distances.  Such  quantities  have 
been  called  "  second  moments  of  mass,"  the  term  being  in  line  with 
"first  moment  of  mass,"  which  is  applied  to  quantities  expressed 

by  integrals  like  /  dm-x  (see  arts.  227  and  228).     We  distinguish 

between  second  moments  of  mass  employing  special  names  for 
the  kinds. 

254.  Moment  of  Inertia. — The  moment  of  inertia  of  a  body 
with  respect  to  any  axis  is  the  sum  of  the  products  obtained  by 
multiplying  each  elementary  mass  of  the  body  by  the  square  of 
its  distance  from  the  axis.  The  axis  will  often  be  called  "inertia- 
axis"  to  distinguish  it  from  other  axes,  coordinate,  geometri- 
cal, etc. 

[Euler  introduced  the  term  "moment  of  inertia,"  and  he  explained  its 
appropriateness  somewhat  as  follows  ("Thoria  Motus  Corporum  Soli- 
dorum,"  p.  167) :  The  choice  of  the  name,  moment  of  inertia  (Ger.  tragheits- 
moment)  is  based  on  analogies  in  the  equations  of  motion  for  translations 
and  rotations.  In  a  translation,  the  acceleration  is  proportional  directly 
to  the  "accelerating  force"  and  inversely  to  the  mass,  or  "inertia,"  of 
the  moving  body,  and  in  a  rotation  the  acceleration  (angular)  is  propor- 
tional directly  to  the  moment  of  the  accelerating  force  and  inversely  to  a 

;       quantity,  jr^dm,  depending  on  the  mass,  or  inertia;    this  quantity,  to 
;      complete  a  similarity,  we  may  call  moment  of  inertia.     Then  we  have — 

for  translations,     acceleration  =  (force)  /(inertia,  or  mass) ; 

for  rotations,  acceleration  =  (moment  of  force)  /(moment  of  inertia)]. 

249 


250  ROTATION.  [Chap.  XIL 

Expression  for  Moment  of  Inertia. — Let  /  *  denote  the  moment 
of  inertia  of  a  body  with  respect  to  an^  axis,  dm  the  mass  of  any 
elementary  portion  all  points  of  which  are  equidistant  from  the 
axis,  and  p  that  distance.     Then  the  definition  states  that 


/dm 


p' (I) 


the  limits  of  integration  being  such  that  all  the  elementary  parts 
of  the  body  are  included  in  the  integration. 

If  the  body  is  homogeneous,  d  denoting  its  density  and  dV 
the  volume  of  the  elementary  mass,  dm  =  ddV\  hence 


=sfdy-p^ .   (2) 


Units  of  Moment  of  Inertia. — From  eq.  (i)  it  is  plain  that 
the  units  involved  in  a  moment  of  inertia  are  those  of  mass 
and  length,  and  hence  the  unit  of  moment  of  inertia  will  de- 
pend upon  the  units  of  mass  and  length  employed.  No  names 
are  in  use  for  the  different  units  of  moment  of  inertia,  but  each 
unit  is  described  by  stating  the  corresponding  units  of  mass  and 
length.  Thus  a  moment  of  inertia  computed  by  using  the  pound 
and  foot  is  said  to  be  expressed  in  a  pound-foot  unit.  The  unit 
corresponding  to  the  geepound  and  foot  may  be  called  the  engi- 
neers' unit  of  moment  of  inertia  or  the  geepound-foot  unit.f 

255.  Radius  of  Gyration. — Since  a  moment  of  inertia  is  one 
dimension  in  mass  and  two  in  length,  it  can  be  expressed  as  the 
product  of  a  mass  and  a  length  squared;  it  is  sometimes  con- 
venient to  so  express  it.  » 

Definition. — The  radius  of  gyration  of  a  body  with  respect  to 
an  axis  is  such  a  length  whose  square  multiplied  by  the  mass  of 
the  body  equals  the  moment  gf  inertia  of  the  body  with  respect 
to  that  axis.  That  is,  if  ^  and  /  denote  the  radius  of  gyration 
and  moment  of  inertia  of  the  body  with  respect  to  any  axis  and 
m  its  mass, 

k2m==I     or     k  =  Vl7m. 

*  A  subscript  affixed  to  the  symbol  refers  to  the  inertia-axis;  thus  Ix 
stands  for  moment  of  inertia  with  respect  to  an  x  axis. 

t  For  dimensions  of  a  unit  moment  of  inertia,  see  Appendix  C. 


§1.] 


SECOND  MOMENTS   OF  MASS. 


251 


The  square  of  the  radius  of  gyration  of  a  homogeneous  body 
with  respect  to  any  axis  is  the-  mean  of  the  squares  of  the  dis- 
tances of  all  the  equal  elementary  parts  of  the  body  from  that 
axis.  For  let  p^,  p^,  etc.,  be  the  distances  from  the  elements, 
dm,  to  the  axis,  and  let  n  denote  their  number  (infinite).  Then 
the  mean  of  the  squares  is 

{pi^+p2^+  .  .  .  )/n  =  (pj^dm+p2^dm+  .  .  .  )/n  dm=I/m, 
But  I/m  is  the  square  of  the  radius  of  gyration,  hence,  etc. 


EXAMPLES. 

1.  Show  that  the  moment  of  inertia  and  radius  of  gyration 
of  a  homogeneous  right  circular  cylinder  with  respect  to  its  geo- 
metric axis  are  respectively 

^mr"^     and     r\/i/2,' 

m  denoting  its  mass  and  r  the  radius  of  its  base. 

Solution :  Let  a  denote  the  altitude  and  imagine  the  cylinder 
to  consist  of  elementary  prisms  parallel  to  the 
axis  and  extending  from  base  to  base.  If  dA 
denotes  the  area  of  the  cross-section  of  any 
element,  then  dV  =  adA,  and  if  p  denotes  the 
distance  of  the  element  from  the  axis, 

•       I  =  dJadA'p\ 

For  convenience,  select  the  prisms  as  indi- 
cated in  fig.  191;  then  dA=pdp-dd,  and 

/  =  adf^  £  "  p^dp .  dO  =  adiTzr^  =  ^mr\ 

Since  k^  =  I/m,  k  =  r\/i/2. 

2.  Show  that  the  moment  of  inertia  and  radius  of  gyration 
of  a  homogeneous  parallelopiped  with  respect  to  one  of  its  geomet- 
rical axes  are  respectively 

w(a2  +  62)/i2     and     V  (a^^+b^j/l2 , 

m  being  the  mass  and  a  and  b  the  lengths  of  the  edges  which  are 
perpendicular  to  the  inertia- axis. 


^5^ 


ROTATION. 


[Chap  XII. 


3.  Show  that  the  moment  of  inertia  and  radius  of  gyration 
of  a  homogeneous  sphere  with  respect, to  a  diameter  are  respec- 
tively 

and     rV2/5. 


mr 


m  being  its  mass  and  r  its  radius. 

Solution:  Let  fig.  192  represent  a  diametrical  section  of  the 
sphere  and  0  Y  the  inertia-axis .  Imagine 
the  sphere  to  consist  of  elementary  1am- 
inas  perpendicular  to  OF;  then  the  mass 
of  the  lamina  is  given  by- 

dm  =  dK{r^  —y^)dy. 

Now  according  to  ex.  i ,  the  moment  of 

inertia  of  the  lamina  with  respect  to  its 

Fig.  192.  geometrical    axis   (OY)  is    ^dm(r^—y^), 

or  -h^7r{r^—y^ydy.     Therefore  the  moment  of  inertia  of  all  the 

laminas,  or  of  the  sphere,  is  the  sum  of  all  such  moments,  i.e., 

/  =  ^d7zf^^\r^  -  y^ydy  =  j\d7:r'  =  tmr\ 

^  4.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of 
a  homogeneous  right  circular  cone  with  respect  to  its  geometrical 
axis  are  respectively 

.2 


mr^     and     rV3/io, 


m  being  its  mass  and  r  the  radius  of  the  base. 

5 .  Show  that  the  moment  of  inertia  of  a  circular  lamina  with 
respect  to  a  diameter  of  either  base  is  approximately  iwr^, 
m  and  r  being  its  mass  and  the  radius  of  the  bases  respectively. 

Solution :  Let  /  denote  the  thickness  of  the  lamina.  Imagine 
it  to  consist  of  elementary  prisms  whose 
bases  coincide  with  those  of  the  lamina, 
their  cross-sections  being  as  shown  in  fig. 
193.  The  volume  of  each  elementary 
prism  is  given  by  tpdd-dp.  Now  all  parts 
of  each  elementary  prism  are  not  equally 
distant  from  the  inertia-axis,  but  they  are  „ 

nearly  so,  except  in  the  case  of  prisms  near 
the  axis.     These  prisms,  however,  contribute  little  to  the  mo- 


§!•] 


SECOND  MOMENTS  OF  MASS. 


253 


ment  of  inertia  of  the  lamina  and  the  error  made  in  assuming  the 
parts  of  such  near  prisms  as  equally  distant  from  the  axis  is 
small.  For  any  prism  the  distance  is  |0  sin  ^;  therefore  approx- 
imately * 

/  =  td£  "  j[  sin^  e  dO '  pHp  =  inrHd. 

Hence,  etc. 
I  ^^     6.  Show  that  the  radius  of  gyration  of  a  thin  elliptic  plate 
with  respect  to  either  of  its  axes  is  ^a,  2a  being  the  length  of  the 
other  axis. 

256.  Relations  Between  Moments  of  Inertia  and  Between 
Radii  of  Gyration  with  Respect  to  Parallel  Axes. — Proposition. — 
The  moment  of  inertia  (/)  with  respect  to  any  axis  equals  the 
moment  of  inertia  (/)  with  respect  to  a  parallel  central  axis  plus 
the  product  of  the  mass  (m)  and  the  square  of  the  distance  (d) 
between  the  axes,  that  is, 

I=I+md' (i) 

Proof:  Let  fig.  194  represent 
a  section  of  the  body  perpen- 
dicular to  the  inertia- axis.  Let  O 
and  C  be  the  points  where  that 
axis  and  the  parallel  central  axis 
respectively  pierce  the  section. 
Then  (see  the  figure)  Fig.  194. 

I=jdm'p^,  and  since  p^=y^-\-{x-\-dy, 

I  =  fdm(y^  +x^  +  2xd  +  d^) 

=  J  dm{x'^-\-y^)-{-2d  idm-x  +  d'^  jdm. 

'Now  J  dm{x^+y^)=  I]   jdm-x  =  mx  =  o\   and  y  Jm=w;   hence, 


etc. 


or 


Corollary:  Dividing  the  sides  of  eq.  (i)  by  m  we  get 
I/m  =  I/m+d'^, 

k2=P-fd2. (t) 


*  The  approximation  is  the  closer  the  less  the  thickness. 


\ 


254  ROTATION.  [Chap.  XII, 

k  denoting  radius  of  gyration  of  a  body  with  respect  to  any  axis, 
k  that  with  respect  to  a  parallel  cefntral  axis,  and  d  the  dis- 
tance between  the  axes. 

Equations  (i)  and  (2)  show  that  the  moment  of  inertia  and 
radius  of  gyration  of  a  body  with  respect  to  a  central  axis  are 
less  than  for  any  other  parallel  axis,  and  that  approximately , 
when  k  is  small  compared  to  d,k  =  d  and  I  =  md? 

EXAMPLES. 

^  I .  Determine  the  moment  of  inertia  of  a  homogeneous  par- 
allelopiped,  the  lengths  of  its  edges  being  a,  6,  and  c  and  its  mass 
tw,  with  respect  to  the  third  edge.  Ans.  m{d^  +  h^)/7,. 

V  2.  Determine  the  radius  of  gyration  of  a  homogeneous  sphere 
with  respect  to  a  line  whose  distance  from  the  centre  is  x. 

,  3.  Determine  the  radius  of  gyration  of  a  right  circular  cylin- 
der with  respect  to  a  line  parallel  to  its  axis  distant  10  ins.  there, 
from,  the  radius  of  the  base  being  4  ins. 

\.  4.  Compute  the  radius  of  gyration  of  a  rod  24  ins.  long  and 
iX  I  in.  in  cross-section  with  respect  to  a  line  parallel  to  a  i-in. 
edge  and  12  ins.  from  the  centre  of  the  rod. 

^^  5 .  Show  that  the  moment  of  inertia  of  a  right  circular  cylinder 
with  respect  to  a  central  axis  parallel  to  the  bases  is  w(rV4  + 
a^/12),  m  denoting  mass,  a  altitude,  and  r  radius  of  the  base. 

Solution:  Imagine  the  cylinder  to  consist  of  elementary  cir- 
cular laminas.  Call  the  distance  of  any  one  of  them  from  the 
inertia-axis  x,  then  its  thickness  is  dx  and  its  volume  is  nr'^dx. 
According  to  ex.  5,  art.  255,  the  moment  of  inertia  of  a  lamina 
with  respect  to  its  central  axis  which  is  parallel  to  the  inertia- 
axis  is  ic/mr^,  and  according  to  eq.  (i),  its  moment  of  inertia 
with  respect  to  the  inertia-axis  of  the  cylinder  is 
\dm  •  r^  +  dm  •  x^  =  \d7:r^dx  +  dnrH'^dx. 

Therefore  the  moment  of  inertia  of  all  the  laminas,  or  of  the 
cylinder,  is  given  by  (if  a  denotes  the  altitude), 

<=23j!r\ir'a  +  ^'-ja^);  etc. 


§1] 


SECOND  MOMENTS  OF  MASS. 


255 


6.  Show  that  the  moment  of  inertia  of  a  right  elliptic  cylinder 
with  respect  to  a  central  axis  parallel  to  either  axis  of  the  base  is ' 
m(aV4  +  ^Vi2),  2a  being  the  length  of  the  other  axis  of  the  base 
and  m  and  h  the  mass  and  altitude  of  the  cylinder  respectively. 
^  7.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of 
a  hollow  right  circular  cylinder  with  respect  to  its  geometrical 

axis  are  respectively   o 

)  

hm{r,^+r^^)     and     \/{r^^  +  r^^)/2, 
m  being  its  mass  and  r^  and  rg  the  inner  and  outer  radii  of  its 
bases. 

257.  Composite  Bodies. — We  refer  now  to  bodies  which  can 
be  divided  into  simple  component  parts;  thus,  a  flywheel  con- 
sists of  hub,  spokes,  and  rim  whose  forms  are  usually  simple. 
The  moment  of  inertia  of  such  a  body  with  respect  to  any  axis 
can  be  computed  by  adding  the  moments  of  inertia  of  the  com- 
ponent parts  taken  with  respect  to  that  same  axis. 

EXAMPLE. 

Compute  the  moment  of  inertia  and  radius  of  gyration  of  the 
cast-iron  flywheel  (weight  450  Ibs.-per-cu.  ft.)  represented  in  fig. 
195,  the  spokes  being  four  in  number  and  elliptic  in  cross-section. 


<— 2^^->k- 


-KH 


Fig.  195. 
258.  Experimental  Determination  of  Moment  of  Inertia. — 

There  are  a  number  of  methods;  we  give  two. 

Pendulum  Methods. — (i)  Suspend  the  body  from  an  axis 
coinciding  with  or  parallel  to  the  inertia-axis.  Let  it  oscillate 
like  a  pendulum  noting  the  '  'time  "  of  an  oscillation  ( T) ,  and  deter- 
mine the  distance  (a)  from  the  axis  of  suspension  to  the  centre  of 
gravity  of  the  body.     Then  substitute  these  values  in  the  equation 

r  =  7r\/F7^*     or     k  =  {T/7r)Vg'/^    ^ 

*  This  is  the  formula  for  the  time  of  oscillation  of  a  pendulum  (see 
art.  267). 


256  ROTATION.  [Chap  XII- 

{g  denoting  the  acceleration  due  to  gravity) ,  and  solve  for  k.  This 
is  the  radius  of  gyration  of  the  body»with  respect  to  the  axis  of 
suspension,  and  from  which  the  methods  of  art.  256  the 
desired  moment  of  inertia  can  be  computed. 
*  (2)  From  the  same  axis  about  which  the  suspended  body 
swings,  suspend  by  means  of  a  cord  a  body  whose  dimensions 
are  small  compared  with  the  length  of  the  cord.  Adjust  the 
length  of  the  cord  so  that  the  times  of  oscillation  of  the  two  sus- 
pended bodies  are  equal.  Then  measure  the  distance  (/)  from 
the  axis  of  suspension  to  the  centre  of  gravity  of  the  small  body 
and  solve  for  k  in  the  equation 

kya^l     or     k  =  Var* 

k  and  a  having  the  same  meanings  as  in  (i). 

Torsion-Balance  Method. — There  are  many  variations  of  the 
method  here  given.  The  balance  for  the  present  purpose  may 
be  arranged  as  follows:  Suspend  an  elastic  wire  vertically,  mak- 
ing the  connection  between  the  wire  and  support  rigid,  and  fasten 
a  flat  plate  in  a  horizontal  position  rigidly  to  the  lower  end  of 
the  wire — the  form  of  the  plate  to  be  such  that  its  moment  of 
inertia  with  respect  to  the  wire  can  be  computed. 

Place  the  body  whose  moment  of  inertia  is  desired  and  a 
second  body  on  the  plate  so  that  they  "  balance,"  i.e.,  leave  the 
plate  horizontal.  The  first  body  is  to  be  so  placed  also  that  the 
wire  is  parallel  to  the  axis  with  respect  to  the  inertia-axis  and 
the  form  of  the  second  body  is  to  be  such  that  its  :pioment  of 
inertia  with  respect  to  the  wire  can  be  computed.  Now,  cause 
the  loaded  balance  to  oscillate  and  note  the  time  (T)  of  one  oscil- 
lation ;  then  remove  the  two  bodies  and  cause  the  empty  balance 
to  oscillate,  noting  the  time  (Tj). 

Let  7i,  /j,  and  /  denote  the  moments  of  inertia  with  respect 
to  the  wire  of  the  plate,  of  the  second  body,  and  of  the  one 
whose  moment  of  inertia  is  desired  respectively;  then,  as  shown 
in  art.  268, 

T:T,::Vl,+I,+I:Vr„ 
or  I^IjyT^'-{I,+h), 

*  Proved  in  art.  267. 


§  11.]  GENERAL  PRINCIPLES.  257 

259.  Product  of  Inertia. — Definition. — The  product  of  inertia 
of  a  body  with  respect  to  two  coordinate  planes  is  the  sum  of 
the  products  obtained  by  multiplying  each  elementary  mass  of 
the  body  by  its  ordinates  from  those  planes. 

Expression  for  Product  of  Inertia. — Let  /  denote  the  product 
of  inertia  of  a  body  with  respect  to  any  two  planes,  as  the  yz 
and  zx  coordinate  planes,  dm  the  mass  of  any  **  third  order  "  ele- 
mentary volume,  and  x,  y,  and  z  its  three  coordinates;  then  the 
definition  states  that  j 


J  =  /dm 


xy, 


the  limits  of  integration  being  so  assigned  that  all  the  elementary 
parts  of  the  body  are  represented  in  thef  integration. 

Since  x  and  y  have  signs,  the  product  dm-xy  may  be  positive 
or  negative,  and  hence  a  product  of  inertia  may  be  (unlike  a 
moment  of  inertia)  negative  or  zero. 

260.  Principal  Axes. — It  can  be  proved  that  through  any 
point  of  a  body  there  are  three  mutually  rectangular  axes  with 
respect  to  two  of  which  the  moments  of  inertia  are  greater  and 
less  than  with  respect  to  any  other  axis  through  the  point.  The 
three  axes  are  called  the  principal  axes  of  the  body  at  that  point. 

The  condition  that  a  line  may  be  a  principal  axis  of  a  body 

at  some  point  of  its  length  is  that  /  dm-xz  and  /  dnt-yz  equal 

zero,  the  line  being  regarded  as  a  2  axis  and  the  point  as  origin. 
(Proof  must  be  omitted.)     It  follows  from  the  foregoing  that 

( 1 )  An  axis  of  symmetry  of  a  homogeneous  body  is  a  princi- 
pal axis  at  every  point  of  it. 

(2)  Any  line  perpendicular  to  a  plane  of  symmetry  of  a  homo- 
geneous body  is  a  principal  axis  at  the  point  where  it  pierces  that 
plane. 

§11.     General  Principles. 

261.  The  Effective  Forces. — Each  particle  of  a  rotating  body 
revolves  in  a  circle  whose  plane  is  perpendicular  to  the  axis  of 
rotation;  therefore  the  acceleration  of,  and  hence  the  effective 
force  for,  each  particle  has  no  component  along  the  axis. 


25^  ROTATION.  [Chap.  XII. 

Let  the  irregular  outline  (fig.  196)  represent  a  rotating 
body,  the  axis  of  rotation  beiiig  perpendicular  to   the  plane 

bf  the  figure  at   0,  and  P  any  par- 
ticle;   also  let 
dm  denote  the  mass  of  P ; 
r  its  distance  from  the  axis ; 
a  "  acceleration; 

a  the   angular    acceleration  of  the 
Fig.  196.  body; 

oj  its  angular  velocity. 
Then  the  effective  force  for  P  is  dm -a,  its  direction  being  the 
same  as  that  of  a  and  the  tangential  and  radial  components  of 
the  force  are  respectively  (since  at  =  ra  and  an  =  rco'^,  art.  216) 

dm-ro:     and     Am-rco^. 

262.  Moment  of  the  Effective  System. — Of  the  two  compo- 
nents of  the  effective  force  for  any  particle,  only  the  tangential 
one  has  a  moment  about  the  axis ;  hence  the  sum  of  the  moments 
of  the  effective  forces  is 


/  {dm 'ra)r  =  a  I  dm -r^  =  la^ 


I  being  the  moment  of  inertia  of  the  body  with  respect  to  the 
axis  of  rotation. 

263.  Equations  of  Motion. — According  to  D'Alembert's  prin- 
ciple, the  sums  of  the  moments  of  the  external  and  effective  forces 
are  equal.  Hence  if  IM  denotes  the  sum  of  the  moments  of  the 
external  forces  about  the  axis  for  any  instant, 

i'M  =  Ia, (i) 

a  being  the  angular  acceleration  of  the  body  at  that  instant . 
This  is  the  equation  of  motion  of  the  body. 

As  shown  in  art.  239  the  sum  of  the  components  of  the  effec- 
tive forces  along  any  line  equals  the  product  of  the  mass  of  the 
body  and  the  component  of  the  acceleration  of  the  mass-centre 
along  that  line.  We  wish  this  sum  for  three  lines,  the  tangent, 
the  normal  to  the  path  of  the  mass-centre  at  the  mass-cen- 
tre, and  the  axis.     These  sums  are  respectively 

mat  =  mra,     mdn  =  mraj^,  ■  and     o, 


§11.]  GENERAL  PRINCIPLES.  259 

wherein  m  denotes  the  mass  of  the  body; 

at  the  tangential  accel  3ration  of  the  mass-centre ; 

a„   "    normal  acceleration  of  the  mass-centre; 

r     "    distance  from  the  mass-centre  to  the  axis. 
If  2T,  IN,  and  I- A  denote  the  algebraic  sums  of  the  compo- 
nents of  the  external  forces  along  the  tangent,  normal  and  axis 
respectively,  then  according  to  D'Alembert's  Principle, 

i'T  =  mra:,     i'N  =  mra;^     i'A  =  o.  .     .     .     .     (2) 

These  equations  may  be  called  the  equations  of  motion  of  the 
mass-centre,  since  they  involve  terms  depending  on  its  motion. 
However,  they  are  useful  not  to  determine  motion  but  especially 
to  determine  forces  when  the  motion  is  known. 

264.  Resultant  of  the  Effective  System. — It  is  assumed  in  this 
article  that  the  rotating  body  is  homogeneous  and  that  it  has  a 
plane  of  symmetry  perpendicular  to  the  axis  of  rotation.  Then 
the  axis  of  rotation  is  a  principal  axis  of  the  body  where  it  pierces 
the  Diane  of  symmetry  (art.  260). 

Imagine  the  body  divided  into  elementary  rods  parallel  to 
the  axis  of  rotation  and  then  one  of  these  rods  into  elementary 
portions  of  equal  length.  These  portions  have  at  any  instant 
the  same  acceleration,  and  hence  the  effective  forces  for  them  are 
equal  and  have  the  same  direction.  It  follows  that  the  resultant 
of  these  effective  forces  is  a  single  force  whose  action  line  is  in  the 
plane  of  symmetry.  The  effective  forces  for  all  the  rods  there- 
fore constitute  a»  coplanar  system,  its  plane  being  the  plane  of 
symmetry.  The  resultant  of  a  coplanar  system  of  forces  is  in 
general  a  force,  and  in  special  cases  a  couple  (arts.  44  and  45); 
we  proceed  to  determine  these. 

Let  fig.  197(a)  represent  the  plane  of  symmetry  of  a  body 
perpendicular  to  the  axis  of  rotation,  0  the  intersection  of  the 
axis  and  the  plane,  and  C  the  mass-centre.  Also  in  addition  to 
the  preceding  notation  let 

k  be  the  radius  of  gyration  of  the  body  with  respect  to  the  axis; 

athe  acceleration  of  the  mass-centre; 

R  denote  the  resultant  effective  force; 

Rt  its  tangential  component  (_L  to  OC) ; 

Rn  "    normal  component  (||  to  0C)\ 


26o  ~  ROTATION.  [Chap.  XIL 


According  to  art.  263  the  sums  of  the  components  of  the  effec- 
tive forces  along  and  perpendicular  to  OC  equal  respectively 
fnan  =  mrio^  and  mat  =  mra.  Therefore  the  components  of  the 
resultant  effective  force  and  the  force  itself  are  given  by 

R<  =  mra:,     R„  =  mfit>2^     R  =  ma, 

and  the  direction  of  R  is  the  same  as  that  of  the  acceleration  of 
the  mass-centre  (a). 

The  action  line  of  the  resultant  cuts  the  line  OC  at  a  point  Q 
whose  distance  from  0  is  k^/r.  This  may  be  proved  as  follows : 
Imagine  th^  resultant  force  to  be  resolved  into  its  two  compo- 
nents {nira  and  mrop)  at  Q.  The  moment  of  the  force  about  O 
equals  {mra)q.  But  the  moment  of  the  force  equals  the  sum 
of  the  moments  of  its  components,  and  this  was  shown  in  art. 
262  to  be  Ia\  hence  mraq  =  Ia,   or 

q  =  k^/f. 

Three  Special  Cases. — (a)  The  angular  velocity  is  constant, 
a  =  o.  The  resultant  effective  force  equals  mrco^,  acts  toward 
the  axis  and  through  the  mass-centre  (see  fig.  1976). 

(b)  The  axis  of  rotation  contains  the  mass-centre,  ?  =  0.  The 
resultant  force  equals  zero,  but  g =» ,  i.e.,  the  resultant  is  a  couple 
and  its  moment  is  la  (see  fig.  197c). 

(c)  The  axis  contains  the  mass-centre  and  the  angular  accel- 
eration is  zero.     The  resultant  vanishes  completely.* 

*  It  can  be  shown  that  the  resultant  of  the  effective  system  is  a  force 
or  a  couple  whenever  the  axis  of  rotation  is  a  principal  axis  of  the  rot  at- 


§  in.]  APPLICATIONS.  261 

265.  Centripetal  and  Centrifugal  Force. — When  the  resultant 
of  the  effective  system  is  a  single  force,  that  of  the  external  sys- 
tem is  also  a  single  force,  and  these  resultants  are  identical; 
hence  if  C  denotes  the  component  of  the  latter  resultant  along 
the  line  joining  the  centres  of  rotation  and  mass, 

C=mfw2=mv2/r. 

The  component  C  of  the  external  forces  acting  on  the  rotat 
ing  body  is  called  centripetal  force  and  the  reaction  correspond- 
ing to  it  is  called  centrifugal  force.  Observe  that  the  first  is 
exerted  on  the  rotating  body  by  other  bodies  and  the  second  hy 
the  rotating  body  on  the  others,  and  that  the  centripetal  force 
acts  from  the  mass-centre  towards  the  axis  of  rotation  and  the 
centrifugal  in  the  opposite  direction. 


§  III.     Applications. 

266.  Determination  of  the  Motion. — If  all  the  external  forces 

which  have  a  moment  about  the  axis  of  rotation  are  given  for 
any  instant  during  the  motion,  the  angular  acceleration  of  the 
body  at  that  instant  can  be  computed  from  the  equation  of 
motion  (art.  263).  If  the  acceleration  can  be  thus  determined 
for  each  instant  and  the  initial  conditions  of  the  motion  (i.e.,  the 
angular  velocity  and  the  position  of  the  body  for  any  one  instant) 
are  also  given,  the  motion  can  be  completely  determined. 


EXAMPLES. 

I.  A  body  whose  moment  of  inertia  is  I  is  made  to  rotate 
about  an  axis  through  the  mass-centre  by  a  constant  force  P 
applied  to  a  cord  wrapped  about  a  cylindrical  portion  of  the 

ing  body  at  some  point  of  the  axis,  and  that,  in  this  case,  all  the  results 
of  art.  264  hold  if  fig.  197  represents  a  section  of  the  rotating  body 
perpendicular  to  the  axis  of  rotation,  O  the  point  where  the  axis  is  prin- 
cipal, and  C  the  projection  of  the  mass-centre  on  the  section. 


262  ROTATION.  [Chap.  XII. 

body,  as  shown  in  fig.  198.     Determine  the  motion,  neglecting 
^^^^^^^  axle  friction  and*  supposing  that  the  angu- 

X         ^^N        lar  velocity  and  the  position  of  the  body  at  a 
/  /      ^        I       certain  instant  are  known. 

/     I     o -ji        \  Solution :  The  external  forces  on  the  body 

\     \^^^       J     are  P ,  its  weight,  and  the  reactions  of  the 
^•^^-^--'''^  supports  on  the  axle.     Of  these  only  P  has 

a  moment  about  the  axis,  hence  the  equa- 
^*  ^^  '  tion  of  motion  becomes 

Pr  =  Ia     or     a=Pr/I (i) 

This  equation  shows  that  the  angular  acceleration  is  constant, 
and  for  a  given  "turning  moment"  (Pr)  applied  to  different 
bodies,  their  angular  accelerations  are  inversely  proportional  to 
their  moments  of  inertia  with  respect  to  the  axes  of  rotation. 

Let  the  position  of  the  body  be  specified  by  means  of  the 
angle  0  which  a  fixed  line  of  it  makes  with  a  fixed  reference  line 
as  in  art.  209,  and  suppose  that  when  P  begins  to  act  the  angle  0 
is  zero  and  the  angular  velocity  {oj)  is  also  zero.    Since  a=dio/dt, 

dco  =  {Pr/I)dt     or     oj  =  {Pr/I)t+C,. 

If  time  is  reckoned  from  the  instant  when  P  begins  to  act,  aj  =  o, 
when  /  =  o.     Substituting  these  values  in  the  last  equation  we  get 

0  =  0 +  Ci,     or     Ci  =  o, 

and  hence  aj  =  {Pr/I)t (2) 

Since  co  =  dd/dt, 

dd  =  {Pr/I)tdt,     or     d  =  h{Pr/I)t^-\-C^. 

Now  d  =  o  when  t  =  o,  and  these  values  substituted  in  the  last 
equation  give 

0  =  0 +  C2,     or     ^2  =  0, 

and  hence  d  =  i{Pr/I)P (3) 

'-'-  2.  Suppose  that  the  body  in  ex.  i  is  a  right  circular  cylinder 
of  cast  iron  (weight  450  Ibs.-per-cu.-ft.)  4  in.  thick  and  2  ft.  in 
diameter,  and  that  P  is  applied  at  the  rim,  its  value  being  6  lbs. 
Determine  the  acceleration.  Ans.  0.82  rad./sec.^ 

3.  Suppose  that  at  a  certain  instant  the  wheel  of  ex.  2  is 
rotating  at  20  rev.-per-sec.  in  the  counter-clockwise  direction. 


§111.]  APPLICATIONS,  263 

What  is  its  velocity  10  sees,  later,  P  acting  upon  the  wheel  dur- 
ing that  time? 

4.  Solve  ex.  i,  supposing  that  a  body  whose  weight  is  T^  is 
suspended  from  the  cord  wrapped  around  the  drum. 

Solution:  Let  T  represent  the  tension  in  the  cord.  This  is 
the  force  replacing  P  of  ex.  i,  and  the  equation  of  motion  for  the 
rotating  body  is 

Tr^Ia .     .     .     (i) 

This  contains  two  unknowns  T  and  a;  to  determine  them  we 
write  next  the  equation  of  motion  for  the  suspended  body.  The 
external  forces  on  it  are  T  and  W,  and  as  its  acceleration  is 
downward  the  resultant  force  on  it  is  down,  i.e.,  W  is  larger 
than  Ti   hence 

W-T  =  (W/g)a .     (2) 

a  being  the  acceleration  of  the  suspended  body.  These  two 
equations  contain  three  unknowns,  T,  a,  and  a,  but  we  have 
the  following  additional  relation : 

a  =  ^« ,     •     (3) 

From  these  three  equations  we  find  that 

Wg  Wg/r 

^     W+mgk'/r'     ^^"^     ""     W+mgk'/r^' 

m  denoting  the  mass  of  the  rotating  body  and  k  its  radius  of 
gyration  with  respect  to  the  axis  of  rotation.  The  equations 
also  determine  the  value  of  the  tension ;  thus 

T  =  W-W'/(W-\-mgk'/r^). 

5.  Suppose  that  a  wheel  rotates  about  a  horizontal  axis  "out 
of  centre"  as  represented  in  fig.  199,  and  that  the  only  external 
forces  on  the  wheel  are  its  own  weight  and 

the  reaction   of  the   supports  •(no   friction).       y^        "'^^^ 
The  angular  velocity  when  C  is  directly  to   /  /^\ 

the  right  of  O  being  given  as   coq  counter-  [         -^^  \q     \ 

clockwise,  determine  the  angular  velocity  of  \        o     ' — / — 

the  wheel  in  any  position.                                         \        ,  / 

Solution:  The  equation  of  motion  is  '^ 

-M^cos  d  =  Ia=Id'd/dt\  ^'''-  '99. 


264 


ROTATION. 


[Chap.  XII. 


and  integration  of  it  gives 

from  which  oj  can  be  computed  for  any  value  of  d. 
7      6.  Solve  ex.  14  art.  245  taking   into    account  the  mass   of 
the  pulley  (then  the  tensions  on  opposite  sides  of  the  pulley  are 
unequal) ,  its  radius  of  gyration  with  respect  to  the  axis  of  rota- 
tion being  k,  its  radius  r,  and  its  weight  W . 

Ans    a  = ^= — ^    

Wk^-^iW^  +  W^y 

7.  Solve  ex.  15  art.  245  taking  into  account  the  mass  of 
the  pulley,  its  radius  of  gyration  with  respect  to  the  axis  being  k, 
its  radius  r,  and  its  weight  W. 

8.  Solve  ex.  16  art.  245  supposing  that  ^  =  V7/9  ft.,  r=i\ 
ft.,  1^=144  lbs.,  and  ^  =  32. 

9.  Fig.  200  represents  a  tub  floating  upside  down.  Two 
cords  are  wrapped  about  the  tub  in  opposite  directions  and  lead 
off  in  parallel  directions  over  pulleys  as  shown,  and  sustain 
hodies  W  and  W.  Discuss  the  motion  of  the  tub  under  the 
influence  of  the  suspended  bodies  and  the  fluid  frictional  resist- 
ance which  assume  to  be  proportional  to  the  velocity. 


flwl 


Fig.  200. 


Solution :  Let  6  denote  the  angular  distance  described  by  the 
tub  in  any  time  t  after  starting,  oj  the  angular  velocity,  and  a 
the  angular  acceleration.  Let  F  denote  the  frictional  resistance 
at  any  instant;  then  since  it  varies  as  the  velocity,  F^cco,  c 
being  a  constant  depending  on  the  liquid,  diameter  of  tub  and 
extent  of  the  wetted  surface.  Let  T  denote  the  tensions  in  the 
cords,  evidently  the  same  at  any  instant,  but  not  constant  in 
time.  Also  let  /  denote  the  moment  of  inertia  of  the  tub  with 
respect  to  the  axis  of  rotation,  2r  its  diameter,  m  and  W  the 
mass  and  weight  of  the  suspended  bodies,  and  a  their  accelerq 
tion. 


§  III.] 


APPLICATIONS. 


265 


The  equations  of  motion  for  each  suspended  body  and  the 
tub  are 

W  —T  =  ma     and     2Tr  —  2Fr  =  Ia. 

Combining  these  with  a  =  ra,  we  get 

{I -\-2mr'^)a-\-2rcaj  =  2Wr, 
and  abbreviating,  we  have  as  the  equation  of  motion 

^w+^-=^ (') 

The  first  integration  of  this  gives 

=By~^~/' (^^ 

C 
B 


(O 


and  the  next 


=  ~U  + 


b'        b) 


(3) 


10.  Plot  a  curve  showing  how  the  angular  velocity  changes, 
taking  A,  B,  and  C  as  2,  5,  and  10  respectively. 

11.  Suppose  that  there  are  no  cords  and  suspended  weights 
and  that  the  tub  is  given  an  angular  velocity  co^.  Discuss  the 
motion  of  the  tub  under  the  influence  of  fluid  friction. 


AnS.    (0  =  COnS 


-bt 


(b  being  equal  to  2 re//). 


267.  Pendulums. — A  body  which  rotates  about  a  horizontal 
axis    under    the    influence    of    its 
weight    and    the    reaction  of   the 
support     is     called     a     compound 
or   physical    pendulum.       Let    fig. 
201  represent  a  section  of  such  a 
pendulum    perpendicular     to    the 
"axis  of  suspension"  and  through 
the  mass-centre  C.     Let  O  be  the 
intersection  of  the  axis  of  suspen- 
sion and  the  section,  and  let 
a  denote  the  distance  OC ; 
k  the  radius  of  gyration  with  re- 
spect to  the  axis  of  sus- 
pension ; 
T  **  time  of  one  oscillation  (from  one  extreme  position  to  the 
other) ; 


266  "-  ROTATION.  [Chap.  XII. 

e  the  angle  XOC; 
6 1   **    maximum  value  of  ^; 
W  ''    weight  of  the  petidulum ; 
m  its  mass. 

We  regard  the  counter-clockwise  direction  as  positive  and  as- 
sume that  the  support  is  frictionless,  or  has  no  moment  about 
the  axis  of  suspension.     Then  the  equation  of  motion  becomes 

'^Wasmd= mk^a = mkWd/dt\ 

or  d^d/dP=-(ag/k^)  ^ind (i) 

The  complete  integration  of  eq.  (i)  is  expressible  by  an  infinite 
series,  but  it  is  not  here  given  because  we  wish  the  value  in  a, 
special  case  which  admits  of  a  simple  approximate  integration. 

We  will  assume  that  the  amplitude  of  the  oscillations  (d^)  is 
so  small  that  practically  sin  d  =  d;   then  eq.  (i)  becomes 

d'd/dt'=-{ag/k')d, (2) 

the  first  integration  of  which  gives 

{dd/dty=  -{ag/k^)d^  +  C^. 

Now  when  6  =  6^,  dd/dt  =  o\  therefore  substituting  these  values 
in  the  last  equation  we  find  that 

o==-{ag/k^)d^  +  C,,     or     C,  =  {ag/k')d,\ 
and  (dd/dt)=±(ag/ky{d^-d')K    ....     (3) 

The  plus  or  minus  sign  is  to  be  used  according  as  dd/dt  is  posi- 
tive or  negative,  i.e.,  according  as  the  pendulum  is  swinging  in  the 
positive  or  negative  direction.     The  integration  of  eq.  (3)  gives 

sin-^  (d/d,)  =  ±(ag/kyt  +  C,. 

Now  if  we  reckon  time  from,  the  instant  when  the  pendulum 
passes  through  its  lowest  position,  i.e.,  ^  =  0  when  ^  =  0,  the  last 
equation  becomes  for  these  values 

sin~^  (o)=o  +  C2;  hence  6*2  =  0. 
If  when  t=o  the  pendulum  is  moving  in  the  positive  direction, 

or  e=e,sm{V^^i)    ) "^ 


§  I".] 


APPLICATIONS. 


267 


This  equation  is  analogous  to  that  for  a  simple  harmonic  motion 
(see  art.  179),  iand  hence  the  motion  of  a  pendulum  is  often 
called  "a  simple  harmonic  oscillation." 

The  time  of  an  oscillation  can  be  found  from  eq.  (4).  Let 
/j  and  ^2  denote  the  values  of  t  when  the  pendulum  is  in  its  high- 
est positions  on  the  right  and  left  respectively.  Then  when 
i  =  tj^,  d  =  di,  and  when  t  =  t2,  it  can  be  shown  that  d=  —0^,  hence 


t^  =  \/k^/ag  sin~^  I  ==i7:Vk^/ag, 
and                        ^2  =  ^k^/ag  sin"  ^(  —  i )  =  j7t\/k^/ag, 
or  T^Wk^ag*       (5) 

This  expression  for  T  being  independent  of  d,  shows  that  the 
time  of  oscillation  of  a  pendulum  is  the  same  for  all  values  of  ^^ , 
provided  that  it  is  so  small  that  sin  0^  practically  equals  0^. 

The  point  Q  (in  OC)  whose  distance  from  O  equals  k^/a 
is  called  the  centre  of  oscillation  and  the  line  through  it  parallel 
to  the  axis  of  suspension  is  called  the  axis  of  oscillation .  The 
following  is  a  simple  geometrical  construction  for  locating  the 
centre  of  oscillation:  Let  O  (fig.  202)  be  the  centre 
of  suspension  and  C  the  mass-centre ;  then  OC  =  a. 
Let  k  denote  the  radius  of  gyration  of  the  pendu- 
lum with  respect  to  a  central  axis  parallel  to  the 
axis  of  suspension  and  lay  off  CK  equal  to  k;  join 
O  and  K  and  draw  KQ  perpendicular  to  0K\  then 
Q  is  the  centre  of  oscillation.     For 

CK^^OC-CQ,  or  CQ  =  kya, 
hence  6Q  =  a  +  kya  =  {a'' ^k^)/a  =  kya 
(seeeq.  (2),  art.  256). 


Fig.  202. 


*  Experimental  Proof  that  the  Masses  of  Bodies  are  Proportional  to  their 
Weights  at  the  Same  Place. — The  expression  for  the  time  of  oscillation  of 
a  pendulum  was  deduced  on  the  assumption  that  mass  is  proportional  to 
weight,  for  we  substituted  g  for  W/m  in  deducing  eq.  i.  If  this  substi- 
tution is  not  made  eq.  (5)  becomes  T=Tik\/m/aW. 

Now  take  two  pendulums,  each  consisting  of  a  sphere  suspended  by 
means  of  a  light  cord,  the  lengths  of  the  cords  being  the  same;  also,  to 
make  the  air  resistances  the  same,  the  spheres  should  be  equal  in  size. 
Next,  compare  their  times  of  oscillation;  it  will  be  found  that  they  are 
equal.     Call  this  time   7',  and  the  weights  and  masses  of  the  pendulums 


268  "-  ROTATION.        '  [Chap.  XII. 

By  length  of  a  compound  pendulum  for  any  given  axis  of  sus- 
pension is  meant  the  distance  from  that  axis  to  the  centre  of 
oscillation.  It  is  shown  in  the  next  paragraph  that  this  is  the 
length  of  an  equivalent  "simple  pendulum." 

A  simple  or  mathematical  pendulum  is  an  ideal  one  consisting 
of  a  particle  suspended  by  means  of  a  massless  cord.  Evidently 
this  is  a  special  form  of  the  physical  pendulum,  and  the  preced- 
ing discussion  applies.  Let  /  equal  the  length  of  the  cord,  then 
for  the  simple  pendulum 

k  =  a  =  l,     or     k^/a  =  lf 
hence  T  =  WI/g (6) 

EXAMPLES. 

I.  A  "seconds  pendulum"  (one  whose  time  of  oscillation  is 
one  second)  is  found  to  be  39.12  in.  long  at  a  certain  place. 
What  is  the  value  of  g  at  that  place  ? 

u  2.  Show  that  the  "axes  of  suspension  and  oscillation  are  in- 
terchangeable," i.e.,  that  the  times  of  oscillation  are  the  same 
whether  a  pendulum  oscillates  about  0  or  about  Q. 

3.  Show  that  on  any  line  OCQ  there  are  four  points  or  axes 
about  which  the  pendulum  will  oscillate  with  the  same  time  T. 

4.  According  to  the  law  of  gravitation  (art.  87)  the  attrac- 
tion of  the  earth  and  hence  the  acceleration  due  to  it  varies  in- 
versely as  the  square  of  the  distance  from  the  earth's  centre. 
Or  if  gj  and  gj  denote  the  accelerations  at  two  points  whose  dis- 
tances from  the  centre  are  r  and  r+e  respectively, 

gi/g2  =  ir  +  ey/r\ 
Show  that  if  T^  and  Tj  are  the  times  of  oscillation  of  a  pendulum 
at  the  two  places  respectively,  approximately 
?  TJTz^i-e/r,     and     e  =  r{i~TJT^), 

Wj  and  Wj,  and  m^  and  m^,  respectively.  Evidently  k  has  the  same 
value  for  the  two  pendulums;  also  o.  Hence  the  equation  above  be- 
comes for  the  two  pendulums 

T=-7zkVmJaWi     and     T  ^nk'VmJaW^ 
or  Wi/w2  =  Wi/T^2- 

This  is  practically  the  method  employed  by  Newton;  he  used  hollow- 
wooden  spheres  containing  gold,  silver,  lead,  glass,  sand,  common  salt, 
wood,  water,  and  wheat. 


§  in.]  APPLICATIONS.  269 

268.  Torsion  Balance. — When  a  torsion  balance  (see  art.  258) 
is  displaced  through  a  small  angle  (the  "pan"  being  rotated 
about  the  wire),  the  moment  of  the  couple  required  to  produce 
the  displacement  is  proportional  to  the  displacement.  Thus  if 
M  and  M'  denote  two  values  of  the  displacing  couple  and  0 
and  d'  the  corresponding  displacements 

M/M'_=^d/d',     or     M  =  Cd, 

C  being  an  abbreviation  for  M'/O'. 

The  wire  exerts  upon  the  pan  in  any  displaced  position  a 
couple  whose  moment  is  equal  but  opposite  in  sign  to  that  of 
the  displacing  couple. 

Imagine  the  pan  displaced  an  amount  6^,  and  then  released; 
it  will  oscillate  under  the  influence  of  the  couple  which  the  wire 
exerts  upon  it.  If  /  denote  the  moment  of  inertia  of  the  pan 
(and  its  contents  if  any)  with  respect  to  the  axis  of  the  wire, 
then  the  equation  of  the  motion  is 

-Cd  =  Ia,     or     dW/dt^-={-C/I)d (i) 

This  is  analogous  to  eq.  (i)  art.  267,  therefore  its  solution  is 
left  to  the  student.  He  should  find  that  the  time  of  oscillation 
{T)  is  given  by 

r  =  7rN/77C.  .     .     .     .     ^     .     (2) 

269.  Conical  Pendulum. — A  conical  pendulum  consists  of  a 
body  suspended  from  a  fixed  point  by  a  cord  and  so  that  it  can 
be  made  to  rotate  about  the  vertical  axis 
through  the  fixed  point  (see  fig.  203).  The 
motion  might  be  caused  and  maintained  by 
means  of  a  vertical  board  rotating  about 
the  axis  and  pressing  laterally  against  the 
suspended  body.  We  wish  to  determine 
the  relation  between  the  angular  velocity 
{oj)  of  the  body  when  constant  and  the 
** height"  Qi)  of  the  pendulum. 

Let  P.  denote  the  tension  in  the  string;  Fig.  203. 

W  the  weight  of  the  body; 
m  its  mass; 
\    R  the  pressure  of  the  board  against  the  body; 
n   the  revolutions  per  unit  time. 


270 


ROTATION. 


[Chap.  XU. 


Then  neglecting  air  resistance,  eqs.  (2)  art.  263,  become 

IT=R  =  o 

IN  =  P  sincj)  =  mroj^  =  ml  sin  ^'aj\       •     , 
J'^=P  cos  ^-1^  =  0 , 


(I) 
(2) 
(3) 


Hence       h=g/(o^  =  g/47i^n^,    and    cos  ^=g//a;^=g//47rV.* 

EXAMPLE. 

Suppose  that  the  weight  of  the  rotating  body  (fig.  203)  is 
10  lbs.,  that  it  makes  100  rev.-per-min.  and  /  is  15  in.  What 
are  the  values  of  h  and  P? 

270.  Weighted  Conical  Pendulum  Governor. — This  consists 
of  three  heavy  bodies,  A,  B,  and  C  (fig.  204),  connected  by  light 

links  as  shown,  the  whole  system 
being  supported  at  D  and  revolv- 
ing about  a  vertical  axis  AD.  We 
wish  to  determine  the  height  {h) 
for  a  given  angular  velocity  (co). 

Let  Pj  denote  the  force  exerted  on 
B  by  BD; 
P3  the  force  exerted  on  Bhy  BA : 
W   "   weights  of  B  and  C; 
Wi  "   weight  of  ^. 

Then  the  forces  exerted  upon  A 

and  B  are  as  shown  in  fig.  204(6). 

Eqs.  (2)  art.  263  for  A  become 

lA  =  2P2  cos  ^-Wi^o,     (i) 

and  for  B, 


Fig.  204. 


IN  =  Pi  sin  ^  +  P2  sin  <^  =  (W/g)l  sin  ^  •  co^, 


Hence 


IA=PiCOs<f) 
h- 


P2  cos  (l)  —  W  =  o. 


(2) 
(3) 


2(W+Wi)g/W(o^ (4) 

EXAMPLE. 

Let  W,  =  10  and  W  =  8  lbs.,  and  draw  a  curve  showing  how  h 
varies  with  the  number  of  rev.-per-min. 

*For  any  deflected  position  of  the  cord,  cos  0<i;  hence  o)>y/^l 
and  n>y/g/l/27t.  If  w<\/g/l,  or  n<\/ g/ljzn,  the  pendulum  will  not 
remain  in  any  deflected  position  however  small.  The  time  of  one  rotation 
at  the  critical  speed,  n=\/g///2;r,  is  the  same  as  the  time  of  one  complete 
vibration  of  the  bob  as  a  simple  pendulum  (see  page  268). 


§nL]  APPLICATIONS.  271 

271.  Kinetic  Reactions. — Definition  (repeated  from  art.  246). 
— By  the  kinetic  reactions  upon  any  body  is  meant  such  com- 
ponents of  the  forces  acting  upon  it  which  depend  upon  its 
acceleration.  The  determination  of  kinetic  reactions  of  rotat- 
ing bodies  is  illustrated  in  the  solution  of  some  of  the  follow- 
ing 

EXAMPLES. 


I.  A  cubical  box,  into  which  a  sphere  just  fits  without  pres- 
sure, is  made  to  rotate  about  a  ver-  W| 
tical  axis,  as  shown  in  fig.  205. 
Determine  the  kinetic  reactions  on 
the  sphere  when  the  angular  ve- 
locity and  acceleration  are  a>  and 
a  respectively. 


I 


Solution:    Let    Ri    denote   the 


ILJx 


% 


pressure  of  the  bottom  of  the  box, 
R^  that  of  the  outer  side,  and  R^  Fig.  205. 

the  third  one.  Evidently  the  latter  acts  as  shown  if  the  accel- 
eration is  counter-clockwise.  Equations  (2)  art.  263  become, 
if  W  and  m  denote  the  weight  and  mass  of  the  sphere  respec- 
tively, and  V  the  velocity  of  the  mass-centre, 

IT=R3  =  inra, 

2N  =  i?2  =  ^ro)'^  =  mlP/r, 

IA=R,-'W  =  o,     or    R,  =  W. 

These  show  that  R2  and  R^  are  entirely  kinetic  and  that  R^  is 
static. 

•  2.  Let  r=  15  ins.,  W=io  lbs.,  and  suppose  that  the  angular 
velocity  increases  every  second  by  2  rev.-per-sec.  Determine 
the  kinetic  reactions  when  the  angular  velocity  is  10  rev.-per- 
min. 

3.  A  body  rests  upon  the  floor  of  a  car  which  moves  in  a 
horizontal  circular  curve  of  radius  r  with  a  constant  speed  v. 
Determine  (i)  the  kinetic  reaction  on  the  body  and  (2)  the 
direction  of  the  resultant  pressure  of  the  body  on  the  floor. 

Ans.  (2)  Inclination  to  the  vertical,  tan~^(z;V^^). 


272  ""  ROTATION.  [Chap.  XIL 

4.  A  body  is  suspended  by  means  of  a  cord  from  the  ceiling 
of  a  car  which  moves  in  a  horizontal  circular  curve  of  radius  r 
with  a  constant  speed  v.  Determine  the  direction  of  the  sus- 
taining cord  and  the  tension  in  it. 

5,  Fig.  206  represents  a  car  on  a  tilted  track.  Suppose  that 
the  track  is  a  horizontal  circular  curve  of  radius  r  and  that  the 

car  moves  with  a  constant  speed  v.  Deter- 
mine the  kinetic  reaction  on  the  car  and 
the  angle  of  tilt  which  makes  the  resul- 
tant of  the  flange  pressures  *  zero. 

Solution :  Imagine  each  wheel  pressure 
resolved  into  three  components,  one  par- 
allel to  the  rails,  one  parallel  to  the  ties, 
and  one  perpendicular  to   the  first  two. 
Call  the  sums  of  these  components  R\ 
R^' ,  and  R'"  respectively,  and  the  resul- 
tant of  R''  and  R'''  R;  also  let  P'  and  P"  denote  the  pulls  at 
the  front  and  rear  of  the  car  respectively,  which  assume  to  be 
practically  parallel. 

Since  the  velocity  of  the  car  is  constant,  a==o  and  (see  eqs. 
(2)  art.  263), 

IT  =  P'-P"-R'  =  o, 

IN==R"  cos  (l>^-R'"  sin  (j)=mv^/r, 

IA=R"'  cos  4>-R"  sin  ^^-1^  =  0. 

These  equations  show  that  R'  =  P'-  P" 

and  that  R  =  {m^v'/r^  +  W^)^. 

If  i^"  (the  sum  of  the  flange  pressures)  equals  zero 

IN  =  i?'"  sin  ^  =  mv^/r    and    I A  =  R"'  cos^-W=o\ 
hence,  combining  tan  (^  =  7;Vgr.t 


*.By  flange  pressure  is  here  meant  the  component  of  the  pressure  on 
a  wheel  parallel  to  a  tie  of  the  track. 

t  This  relation  makes  the  sum  of  the  flange  pressures,  but  not  each  one 
necessarily,  equal  to  zero.  It  has  been  discovered  experimentally  that  if 
the  wheels  are  coupled  together  in  fours  as  usual,  the  front  outer  wheel 
always  experiences  a  flange  pressure. 


III.] 


APPLICATIONS. 


273 


These  results  can  be  reached  graphically  a  little  more  simply 
as  follows:  Since  R' ,  P' ,  and  P"  are  in  equilibrium,  R  and  W  are 
equivalent  to  the  effective  force  for  the  car,  i.e.,  the  resultant 
of  R  and  W  is  identical  with  the  resultant  effective  force.  So 
draw  from  C  a  line  Cc  to  represent  the  resultant  effective  force 
mroj^  =  niv'^/r,  a  line  Cw  to  represent  W^  and  complete  the  par- 
allelogram Cwcr.  Then  Cr  represents  R,  and  from  the  figure  it 
is  seen  that  R  equals  the  value  given  in  the  foregoing.  In  order 
that  R  may  have  no  component  along  the  tie,  i.e.,  R"  =  o^ 
the  angle  Crc  must  equal  0,  or 

tan  ^  =  Cc/cr  =  {mv^/r)/W, 

272.  Weight  of  a  Body  as  Influenced  by  the  Earth* s  Rotation. — 
Let  fig.  207  represent  a  meridional  section  of  the  earth,  ON  being 
the  polar  axis.  Imagine  a  body  resting  on 
the  surface  at  A  or  suspended  by  means 
of  a  cord.  The  forces  acting  upon  the 
body  are  two  in  number,  the  attraction  of 
the  earth  (P) ,  and  the  reaction  of  the  sup- 
port or  the  pull  of  the  cord  (Q).  P  is 
directed  somewhat  as  shown;  let  its  mag- 
nitude be  represented  hy  AB.  Q  is  not 
collinear  with  P  (except  at  the  equator  or 
pole)  because  the  resultant  of  P  and  Q 
must  be  directed  the  same  as  the  resultant 
effective  force  for  the  body;  the  direction 
of  this  is  AD  fthe  radius  of  the  path  oi  A). 

Let  R  denote  the  resultant  effective  force  (also  the  resultant 
of  P  and  Q),  m  the  mass  of  the  body,  a>  the  angular  velocity  of 
the  earth,  and  r  the  radius  of  the  path  of  .4 ;  then 

R  =  mr(o^, 

and  if  AD  represents  R,  the  side  ^C  of  the  parallelogram  drawn 
on  ABD  represents  Q.     It  follows  from  the  figure  that 


Fig. 


207. 


or, 


P^  =  Q^+R^  +  2QR  cos  (j), 
Q  =  P\/i-sin2  ct>{R/Py-R  cos  <1>. 


(I) 


2  74  ROTATION.  [Chap.  XII. 

It  is  shown  in  ex.  i  below  that  R/Pis  less  than  1/289;  hence, 
approximately  , 

Q  =  P—R  cos  (l)  =  P—mroji  cos  (j) (2) 

Q  or  its  opposite  is  the  force  which  we  actually  measure  by 
spring-  or  beam-balance  and  call  the  weight  of  the  body.  It 
may  be  called  "apparent  weight"  to  distinguish  it  from  the 
attraction  P,  or  "real  weight."  Eq.  (2)  shows  that  the  apparent 
weight  is  always  less  than  the  real,  and  that  the  difference  de- 
pends on  r  cos  0. 

Notice  that  <j)  is  the  latitude  at  A ,  since  it  is  the  angle  made 
by  the  plumb  line  at  A  with  the  equatorial  plane.  Since  R 
(==mrco^)  is  equal  to  the  centrifugal  force  of  the  body,  the  rela- 
tionship in  eq.  (2)  is  sometimes  expressed  thus:  "The  weight 
of  a  body  is  diminished  by  the  product  of  its  centrifugal  force 
and  the  cosine  of  the  latitude." 

EXAMPLES. 

1.  Show  that  at  the  equator  the  difference  between  the  real 
and  apparent  weights  of  a  body  is  about  1/289  of  the  apparent 
weight. 

Solution:  From  eq.  (i),  since  sin  ^  =  0  at  the  equator 

(P-Q)/Q  =  R/Q  =  ntr,coyQ, 

Tq  denoting  the  equatorial  radius.  Now  ni/Q  =  i/g,  g  denoting 
the  acceleration  due  to  gravity  at  the  equator  as  measured  ex- 
perimentally, i.e.,  g  is  also  "apparent";   hence 

{P-Q)/Q  =  r,a>yg. 

Now  oj  =  27:/t,  where  /  denotes  the  time  of  one  revolution  of  the 
earth;  and  since  /  =  86,i64  sec,  ro  =  20,920,000  ft.,  and  0^  =  32.09 
ft. /sec. ^,  r(ja>Vg  =  0.003467  =  1/289. 

2.  Show  that  if  the  earth  rotated  17  times  as  fast  as  it  does» 
then  the  apparent  weight  of  a  body  at  the  equator  would  be 
practically  zero. 

273.  Centrifugal  Hoop  Tension. — Imagine  a  hoop  to  lie  upon 
a  horizontal  table  which  rotates  about  a  vertical  axis  through 
the  centre  of  the  hoop.     The  tension  which  exists  at  each  cross 
section  of  the  hoop  is  called  "centrifugal  hoop  tension";    we 
now  deduce  an  expression  for  it. 


§  ni.] 


APPLICATIONS. 


275 


Fig.  20 


Let  fig.  208  represent  one  half  of  the  hoop.  The  forces  act- 
ing on  this  half  consist  of  its  weight,  the  reac- 
tion of  the  table,  and  the  forces  exerted  by  the 
other  half,  i.e.,  the  hoop  tensions  at  the  two 
sections.  Since  the  first  two  forces  balance 
each  other,  the  resultant  of  the  remaining  two 
must  equal  the  resultant  effective  force.  Hence 
if  m  denotes  the  mass  of  one  half  of  the  hoop, 
r  the  distance  from  the  axis  to  its  mass-centre, 
oj  its  angular  velocity,  and  P  the  hoop  tension, 
2P  =  mroj^,     or     P=-^mraj^. 

EXAMPLE. 

Let  r  denote  the  radius  of  the  hoop  and  w  its  specific  weight. 
Regard  the  tension  at  a  section  as  uniform  (practically  true 
when  the  thickness  is  small  compared  to  the  radius),  and  show 
that  the  intensity  of  the  hoop  tension  equals  wr^o//g. 

274.  Hinge  Reactions. — Rotating  bodies  often  turn  (a)  about 
a  fixed  shaft  or  (b)  with  a  shaft  in  fixed  bearings.  The  force 
exerted  by  the  shaft  on  the  body  in  the  first  case  and  those 
exerted  by  the  bearings  on  the  shaft  in  the  second  will  be  called 
hinge  reactions.  Determination  of  hinge  reactions  in  the  fol- 
lowing is  limited  to  cases  in  which  the  rotating  body  has  a  plane 
of  symmetry  perpendicular  to  the  axis.  Then  the  resultant  of 
the  effective  system  for  the  body  consists  of  a  single  force  (see 
art.  264). 

Case  I.  Rotation  about  a  Fixed  Shaft. — We  assume  that  the 
applied  forces  are  such  that  the  hinge  reaction  is  equivalent  to 
a  single  force  which  call  R.     Let  fig.  209  represent  a  section  of 

the  rotating  body  through  the 
mass-centre  (C)  and  perpendic- 
ular to  the  axis  (O).  Imagine 
R  resolved  into  three  compo- 
nents, one  parallel  to  OC*  one 
parallel  to  the  axis,  and  one  per- 
pendicular to  OC  and  the  axis, 
and  denote  them  by  Rn,  Ra  (not  shown),  and  Rt  respectively. 

*  If  the  mass-centre  is  in  the  axis  the  direction  of  OC  may  be  taken 
any  way  perpendicular  to  the  axis. 


Fig.  209. 


2  76 


ROTATION. 


[Chap.  XII. 


Let  IFn,  ^Fay  and  IFt  denote  the  algebraic  sums  of  the  com- 
ponents of  all  applied  forces  (^'.^.,  all  the  external  forces  except  the 
hinge  reaction),  parallel  to  Rn,  Ra,  and  Rt  respectively.  The 
components  of  the  resultant  effective  forces  parallel  to  these 
same  directions  are  mrcu^,  o,  and  mray  acting  as  shown  in  fig.  209. 
Since  the  external  forces  and  the  reversed  effective  forces  are 
in  equilibrium  (art.  238), 

or     Rt  =  mra  —  J  Ft, 
or    Rn  =  nirw^  —  IFny 
or    Ra=—  2  Fa. 
These  equations  show  that  Ra  has  no  kinetic  component,  and 
that  the  kinetic   components  of  Rt  and  Rn  equal  zero  if  r  =  o, 
i.e.,  if  the  mass-centre  is  in  the  axis  and  if,  as  was  assumed  at 
the  outset,  the  body  has  a  plane  of  symmetry  perpendicular 
to  the  axis. 

Case  II.  The  Body  Rotates  about  a  Shaft  in  Bearings. — We 

assume  that  there  are  two  bear- 
ings whose  reactions  call  R'  and 
R".  Let  A  and  B  (fig.  210)  be 
the  bearings,  and  the  parallelo- 
gram the  plane  of  symmetry  of 
the  body. 

Imagine  R'  and  R"  (like  R, 
Case  I)  resolved  into  three  com- 
ponents, and   extend  the  nota- 
tion of  that  case  to  the  present 
one.     Also   let    IMt  and   IMn 
denote  the  moment  sums  of  the 
applied   forces   {all   the   external 
forces  not  including  R'  and  R")  with  respect  to  the  lines  marked 
Ot  and  On  respectively.     Then  as  all  the  external  forces  and 
the  reversed  resultant  effective  force  are  in  equilibrium, 
R/+Rt"  +  IFt  =  mra, 
Rn'+Rn'  +  ^Fn  =  mraj\ 

Ra'+Ra''  +  ^F,  =  0, 

Rn"V'-Rv!l'-\-IMt  =  o, 
Rt'V-Rt"V'  +  IMn  =  o. 


Fig.  210. 


These  follow  also  from  eqs.  (2),  art.  263. 


^III.]  ■    APPLICATIONS.  277 

From  these  equations  we  find  that 

R/l  =  {mra  -  IFtW  -  ^Mn, 
Rn'l  =  (wra>2  -  IFn)l"  +  ^Mu 
R/n  =  {m?a  -  IFt)V  +  IMn. 
.      Rn''l=-(m?co^-IFn)l'-IMty 
R  f^R  " ^  -IF  . 

These  equations  show  that  the  kinetic  components  of  the  hinge 
reactions  are  zero  if  r  =0,  i.e.,  if  the  mass-centre  is  in  the  axis, 
and  if,  as  was  assumed  at  the  outset,  the  rotating  body  has  a 
plane  of  symmetry  perpendicular  to  the  axis  of  rotation.  An 
axis  of  a  body  for  which  the  kinetic  components  of  the  hinge 
reactions  are  zero  is  called  a  "free  axis."  It  can  be  shown  that 
the  three  central  principal  axes  of  any  body  are  free  axes. 

EXAMPLES.* 

I.  Suppose  that  in  ex.  5  art.  266,  W=ioo  lbs.,  /=io  (gee- 
pound-foot  units),  ?  =  i/2  ft.,  and  coq  =  4  rad.-per-sec.  Compute 
the  hinge  reaction  when  C  is  directly  to  the  right  of  O. 

Solution:  According  to  the  solution  of  ex.         ^ — 
5,  the  angular  acceleration  in  the  position  un-      /^       j         \. 
der  consideration  is    —  5  rad.-per-sec. -per-sec.   (  ^    S 
Hence  (see  fig.  211)  r — ^ 

i?f -100= —(100/32. 2)-^-5,  or  Rt  =  g2.2  lbs.     \ 
i?n  =  (ioo/32.2)-^- 16  =  24.8  lbs.  \, 

I  2.  Determine  the  hinge  reactions  in  ex.  i,  •^^^-  ^^'^^ 

when  C  is  vertically  above  0,  below  O,  and  to  the  left  of  O. 
V,    3.  Suppose  that  the  wheel  of  the  preceding  example  revolves 
about  a  vertical  axis  with  a  constant  angular  velocity  of  4  rad.- 
per-sec.     Determine  the  hinge  reaction  in  any  position  of  the 
wheel,  AO  and  OB  (see  fig.  210)  being  5  inches.      -^^^   -.  ;:  c   - 

*  The  student  is  advised  not  to  use  the  foregoing  formulas,  but  to 
proceed  as  follows:  (i)  Determine  the  resultant  effective  force  for  the 
rotating  body  in  the  position  under  consideration,  remembering  that  the 
normal  component  acts  from  the  mass-centre  toward  the  axis  and  the 
tangential  component  in  the  direction  of  the  tangential  component  of 
the  acceleration  of  the  mass-centre.  (2)  Write  the  conditions  of  equi- 
librium for  the  force  system  consisting  of  that  resultant  reversed  and  all 
the  external  forces  (including  the  hinge  reactions) .      (3)   Solve. 


278 


ROTATION. 


[Chap.  XII. 


4.  Determine  the  hinge  reaction  on  the  rotating  body  de- 
scribed in  ex.  2,  art.  266.      "^i  ^  P     a>  ■-  - 

5.  Imagine  the  rotating  body  of*fig.  210  to  be  a  parallelo- 

piped  whose  weight  is  90  lbs.,  and 
whose  radius  of  gyration  with 
respect  to  a  central  axis  parallel 
to  the  axis  of  rotation  is  7.84  in., 
that  /'  =  /"  =  18  in.,  and  OC=i  in. 
Suppose  that  the  body  is  rotated 
by  means  of  a  couple  applied  to  a 
thin  disk  (mass  negligible),  as 
shown  in  fig.  212.  Determine  the 
hinge  reactions  when  the  motion  is 
about  to  begin  {10  =  0). 

Solution:  The  masses  of  the 
suspended  and  rotating  bodies  are 
respectively 

90/3  2.2  =  2.79  geepounds. 

The  moment  of  inertia  of  the  rotating  body  about  the  axis  of 
rotation  is 


Fig.  212. 


48/32.2  =  1.49     and 


2.79  (7.84/12)2  +  2.79  (1/12)2  =  1.21  (geepound-foot  units). 

Hence  the  equations  of  motion  of  the  suspended  and  rotating 
bodies  are  respectively,  T  denoting  tension  in  the  cord, 


48  — r=i.49a,     T- 


1.21a 


and  since  here  a  =  a/2,  we  find  from  the  equations  that  a.-^  tj^.^ 
rad.-per-sec.-per-sec. 

The  components  of  the  resultant  effective  force  are 

w7a  =  2.79  XtV"X  24.5  =  5-68     and     o, 

and  act  as  shown  in  the  figure.  Since  the  reversed  resultant 
effective  force  and  the  external  forces  are  in  equilibrium,  we 
next  write  as  many  conditions  of  equilibrium  for  that  system 
as  are  necessary  to  determine  the  unknowns.  Thus,  for  the 
action  line  of  R/'  as  moment  axis. 


i^n'- 3— 90-1/12  =  0,     or    i?„'  =  2.5  lbs., 


§  in.] 


APPLICATIONS. 


79 


for  the  action  line  of  Rn"  as  moment  axis, 

7?/. 3-5. 68-14  =  0     or     i^/  =  2.84lbs.; 
and  since  i?/+i^/'  =  5.68  and  i?„'+i?n"  =  o, 

i^/'  =  2.84     and     7^,/'= -2.5  lbs. 
From  the  "  axis  resolution  equation" 

RJ'-go  =  o,     or     /?/'  =  9olbs. 

6.  Determine  the  hinge  reactions  when  the  body  has  rotated 
through  90°,  180°,  270°,  and  360",  and  record  your  results  in  a 
tabular  form  as  follows: 


a 

(J) 

nira 

mroj^ 

R/ 

Rn' 

Rt" 

J?  " 

Kn 

Ra'' 

0° 

90° 

24.5 
24.5 

0 

5.68 

0 

2.84 

2-5 

2.84 

-2.5 

90 

w  7.  Suppose  that  in  ex.  5  there  is  no  couple  applied  and  that 
the  body  rotates  with  a  constant  speed  of  100  rev.-per-min. 
Determine  the  hinge  reactions. 

8.  Suppose  that  in  ex.  7  there  is  a  second  parallelepiped 
just  like  the  first  attached  to  the  shaft,  also  i  in.  "out  of  cen- 
tre," and  so  that  the  two  mass-centres  are  on  opposite  sides  of 
the  shaft,  the  second  oile  being  3  in.  vertically  above  the  first. 
When  they  are  rotating  at  100  rev.-per-min.,  determine  the 
hinge  reactions. 

27$.  Balancing  of  Rotating  Bodies. — As  illustrated  by  exs. 
5-8  of  the  preceding  article,  when  one  or  more  bodies  rotate 
with  a  chaft,  the  hinge  reactions  have  in  general  kinetic  com- 
ponents, i.e.,  the  hinge  reactions  depend  in  part  at  least  upon 
the  motion.  These  kinetic  components  continually  change 
direction  so  that  the  bearings  are  subjected  to  injurious  influences 
to  prevent  which  it  is  sometimes  desirable  to  "balance"  the 
rotating  system. 

Balancing  consists  in  arranging  the  rotating  bodies  on  the 
shaft  so  that  the  effective  forces  for  the  rotating  system  are  in 
equilibrium;  then  the  kinetic  components  of  the  hinge  reac- 
tions vanish,  and  the  hinge  reactions  depend  only  on  external 
forces. 


28c  -  ROTATION.  [Chap.  XII. 

In  the  following  we  consider  only  rotation  at  constant  speed 
(a  =  o),  and  rotating  systems  consisting  of  one  or  more  bodies 
which  have  planes  of  symmetry  perpendicular  to  the  axis.  Then 
the  resultant  effective  force  for  each  rotating  body  is  a  force 
whose  action  line  is  in  that  plane  and  passes  through  the  mass- 
centre  and  the  axis,  and  whose  magnitude  is  mroj^  {m  denoting 
mass  of  the  body,  r  the  distance  of  its  mass-centre  from  the  axis, 
and  oj  the  angular  velocity  of  the  system). 

Since  the  centrifugal  force  which  each  rotating  body  exeits 
upon  the  shaft  is  equal  and  opposite  to  the  resultant  effective 
force  for  the  body,  balance  is  effected  if  the  centrifugal  forces 
(or  the  reversed  effective  forces)  for  all  the  rotating  bodies  are 
in  equilibrium.  This  is  the  usual  view  of  balancing  and  we  shall 
follow  it. 

For  convenience  we  will  write  r  instead  of  r  and  will  often 
say  "  body"  instead  of  "  mass-centre  of  a  body."  We  will  also 
denote  both  a  body  and  its  mass  by  m. 

276.    Balancing  Bodies  whose  Mass-centres  rotate  in  the  Same 

Plane. — Let  m^,  Wo,  and  m^  (fig. 
213)  represent  three  bodies  whose 
common  centre  of  rotation  is  O. 
^^  Their  centrifugal  forces  equal 
respectively 

m^r^oj^,     m^r^cx)^,     m^r^uj^y 

'  ^^^'  their  directions  being  from  O  to 

the  mass-centre  of  the  rotating  body  in  each  case. 

Let  AB,  BC,  and  CD  represent  the  magnitudes  and  directions 
of  the  three  centrifugal  forces.  Then  a  fourth  force  acting 
through  O,  which  would  close  the  force  polygon  ABCD,  would 
balance  the  three  centrifugal  forces.  This  fourth  or  balancing 
force  can  be  supplied  by  adding  to  the  system  of  rotating  bodies 
a  fourth  one  whose  centre  of  rotation  is  0,  whose  mass-centre  is 
in  the  direction  DA  from  0,  and  whose  distance  from  the  axis 
(r)  and  mass  (m)  are  such  that 

wra>^  =  Z).4  (by  scale),     or     mr  =  DA/oj^. 
It  will  be  noticed  that  a;  is  a  common  term  in  the  forces  rep- 


§  III.]  APPLICATIONS,  281 

resented  by  the  lines  of  the  force  polygon.  Hence,  if  the  poly- 
gon for  the  centrifugal  forces  closes,  it  will  also  close  for  forces 
parallel  to  them  of  magnitudes  m^r^,  Wjrj,  etc.;  so  that  \i  ABy 
BC,  and  CD  are  drawn  parallel  to  the  first  three  centrifugal 
forces  respectively  but  equal  (by  scale)  to  w^rj,  ^2^3,  m^r^,  then 
DA  represents  mr  to  that  same  scale. 

277.  Balancing  Bodies  whose  Mass-centres  Rotate  in  Different 
Planes. — In  general  such  bodies  cannot  be  balanced  by  adding 
to  the  rotating  system  a  single  body.  It  will  now  be  shown 
that  any  rotating  system  can  be  balanced  by  adding  two  bodies 
which  rotate  in  any  selected  planes. 

A  Single  Rotating  Body. — Let  m  denote  the  mass  of  the 
given  body,  r  the  distance  from  its  mass-centre  to  the  axis,  co 
the  angular  velocity,  and  O  (fig.  214a)  its  centre  of  rotation. 


1^- 


4m  {  4m 

1 bI  _£ 


't^ -ir       p^ 

far  ^u  ^ 

Fig.  214a. 

Also  let  A  and  B  be  the  selected  centres  of  rotation  of  the  two 
balancing  bodies,  m'  and  m"  their  masses,  and  r'  and  r"  the 
distances  from  their  mass-centres  to  the  axis  as  shown. 

The  centrifugal  forces  of  m,  w',  and  m"  are  mro?,  m'r'u?^ 
and  m"r" u?  respectively.     In  order  that  they  may  be  in  equi- 
librium, 
(i)  The   plane   of  the  mass-centres  of  the  three   bodies   must 

contain  the  axis  of  rotation,  and 
(2)  The    algebraic    sum   of   the   moments  of  the  three    forces 

about  any  point  in  their  plane  must  equal  zero ;  hence 

m'r'  =  mrhlc     and     m"r"  =  mra/c. 

(Notice  that  it  is  not  necessary  to  use  units  of  a  kinetic 
system  in  these  equations ;  any  unit  of  length  and  unit  of  mass 
may  be  used.  Generally  the  foot  and  pound,  or  inch  and 
pound,  will  be  most  convenient.) 


2^2  "  ROTATION.  [Chap.  XII. 

From  these  two  equations  we  can  compute  m'r'  and  m"r"  in 
terms  of  the  given  quantities,  m,  r,  a,  b,  and  c;  we  may  arbi- 
trarily select  either  of  the  factors  in  w'V  and  m^ V"  and  then  com  - 
pute  the  remaining  two. 

Two  cases  may  be  distinguished:  (a)  the  balancing  bodies 
are  on  opposite  sides  of  the  given  body;  (6)  they  are  on  the 
same  side  of  it.  Condition  (2)  requires  in  each  case  that  the 
middle  one  of  the  three  bodies  (w,  m',  and  m")  shall  be  alone 
upon  one  side  of  the  axis  and  the  outer  ones  together  on  the 
opposite  side  of  the  axis. 

Any  Number  of  Rotating  Bodies. — Let  m^,  Wj,  m^,  etc., 
denote  the  masses  of  the  given  rotating  bodies,  and  let  the 
centres  of  rotation  of  the  balancing  bodies  be  denoted  by  A  and  B. 
By  the  method  just  explained,  determine  two  bodies,  m^'  and  m/' 
rotating  about  A  and  B  respectively  which  will  balance  m^; 
likewise  two  bodies  W2'  and  m^'  rotating  about  A  and  B  re- 
spectively which  will  balance  ^2,  etc.  Then  the  given  bodies 
together  with  m/,  m^^  .  .  .  m/',  mj",  .  .  .  would  be  "in balance." 

Next  compound  the  centrifugal  forces  of  w/,  mj',  .  .  .  and 
determine  a  single  body  m'  whose  centrifugal  force  is  equiva- 
lent to  that  resultant;  likewise  compound  the  centrifugal 
forces  of  w/',  m/',  .  .  .  and  determine  a  single  body  w"  whose 
centrifugal  force  is  equivalent  to  that  resultant.  Then  m' 
and  m''  are  the  two  bodies  sought. 

EXAMPLES. 

I.  In  the  left-hand  part  of  fig.  214(a),  let  w  =  20  lbs., 
r=i^^  a  =  6,  and  6  =  14  in.  Determine  the  balancing  bodies 
w'  and  m'\ 

Solution:   According  to  the  equation  under  (2),  page  281, 

mV  =  (20 'IS*  14)720  =  210  lb. -in. 
and 

m'V"  =  (20-is-6)/2o=   90  lb. -in.* 

If  r'  and  r"  are  taken  as  6  and  8  in.  respectively,  then 
w'  =  2io/6  =  35     and     w"  =  9o/8  =  11.25  lbs. 

*  It  should  be  noticed  that  products  like  mr  are  moments  of  mass, 
see  arts.  227  and  228. 


§  ni.] 


APPLICATIONS. 


283 


^  2.  Solve  the  preceding  example,  but  let  the  data  refer  to 
the  right-hand  part  of  fig.  214  (a). 

3.  Let  mi,  w„  and  m^  (fig.  2146)  be  three  bodies  on  the 


k— 6^—H< -8^- 


Potygon  for    / 

forces  at  A  / 

/ 
/ 

)^ 

I 


—5^ 


I M  ¥\J  \ 


Fig.  214/J. 

shaft  AB,  and  determine  two  bodies  rotating  about  A  and  B 
which  will  balance  them. 

Solution:  The  data  are  arranged  in  the  first  five  columns 

f  of  the  adjoining  tabulation,  6  denoting  the  angles  which  the 
"radii"  of  the  different  rotating  bodies  make  with  the  vertical 
plane  through  the  axis  of  rotation,  and  a  and  b  having  the 

I  same  meanings  as  in  fig.  214  (a).  The  quantities  in  columns 
6  to  12  were  determined  by  the  methods  explained  in  the  fore- 
going text. 


: 

2 

3 

4 

s 

6 

7 

8 

Q 

10 

11 

12 

m 

r 

e 

a 

b 

ntr 

tnra 

tnrb 

m'r' 

ff' 

m"r" 

5" 

10 

6 

0 

5 

15 

60 

300 

900 

45 

180 

15 

180 

8 

7 

240 

13 

7 

56 

728 

392 

19.6 

60 

36.4 

60 

9 

4 

no 

24 

4 

36 

864 

144 

7.2 

no 

43-2 

290 

The  centrifugal  forces  of  the  imagined  balancing  bodies 
rotating  about  A  are  450;^,  19.60;^,  and  7.20;^;  a'h'c'd'  is  their 
force  polygon,  and  hence  a'd'  represents  their  resultant.  The 
angle  which  a'd'  makes  with  the  vertical  is  147°  40',  and  the  ma  s- 
moment  represented  by  a'd'  is  44.5  lb. -in. ;  hence  a  single  body  of 
8.9  lbs.  fixed  to  the  shaft  with  a  radius  of  5  in.  in  the  position 
m'  is  one  of  the  required  balancing  bodies. 

The    centrifugal   force;   of   the   imagined   balancing   bodies 


284 


ROTATION. 


[Chap.  XII. 


rotating  about  B  are  150;^,  36.4<w2,  and  43.20;^;  a"h"c"d''  is 
their  force  polygon,  and  hence  a"d"  represent  their  resul;ant. 
The  angle  which  a^'d"  makes  with  Vhe  vjitical  is  328°  40',  and 
the  mass-moment  represented  by  a"d"  is  20.4  lb. -in.;  hence  a 
single  body  of  3.4  lbs.  fixed  to  the  shaft  with  a  radius  of  6  ins. 
in  the  position  m"  is  the  other  one  of  the  required  balancing 
bodies. 

4.  Move  A  of  fig.  214  (6)  to  a  point  midway  between  0^ 
and  O2,  and  move  S  to  a  point  4  in.  to  the  right  of  0^\  then  solve 
the  preceding  example. 

278.  Pivot  and  Journal  Friction. — Flat  Pivot. — Let  fig.  215(a) 
represent  the  flat  end  of  a  shaft  which  is  pressed  or  "thrust" 
against  a  flat  bearing  and  rotates.  Let  r  denote  the  radius 
of  the  end,  N  the  thrust  or  normal  pressure,  and  /  the  coeffi- 
cient of  friction  for  the  rubbing  surfaces. 

Regarding  the  normal  pressure  as  uniform,  its  value  per  unit 
area  is  N/nr'^,  and  the  normal  pressure  on  any  elementary 
area  as  that  indicated  in  fig.  215(a)  is  (N / nr^) pdO dp  while  the 
friction  on  that  area  is  /  times  that  elementary  pressure.  The 
moment   of  the   elementary  frictional  force  about  the  axis  of 

In  4p 


(a)        -  C6) 

Fig.  215. 
the  shaft  is  p  times  the  force,  and  the  sum  of  all  such  moments 
Is 


f(N/7:r')£  J"'p'dpde  =  fNii 


The  resultant  friction  is  not  a  force  but  a  couple,  and  hence  we  ; 
may  regard  the  actual  frictional  resistance  as  a  couple  whose  l 
forces  equal  fN  and  whose  arm  is  |r.  J 


g  III.]  JPPUC^TIONS.  285 

Conical  Pivot. — Let  fig.  215(6)  represent  a  rotating  conical 
pivot  which  is  pressed  into  its  step  by  a  thrust  P  directed  along 
the  axis.  let  r  denote  the  radius  of  the  step,  cj)  the  angle  shown, 
and  p  the  normal  pressure  per  unit  area  regarded  as  constant. 

The  normal  pressure  on  an  elementary  area  dA  of  the  bear- 
ing is  pdA  and  its  vertical  component  is  (pdA)  sin  ^.  Since 
the  friction  has  no  vertical  component  the  sum  of  all  the  ver- 
tical components  of  the  normal  pressure  must  equal  P,  hence 

P  =  p  sin  (f)fdA  =p  sin  ^'A=  pnr^ ,     or     /?  =  P/nr^. 

Note  that  the  normal  pressure  per  unit  area  is  independent  of  ^. 
The  frictional  force  on  each  element  of  area  dA  is  j{P/nr'^)dA , 
and  its  moment  with  respect  to  the'axis  is  p  times  the  force  (see 
the  figure) ;  hence  the  entire  frictional  moment  equals 

KP/Kr')fdA-p. 

For  simplicity  take  dA  of  such  form  that  its  horizontal  projec- 
tion equals  pdpdd  (see  the  figure),  i.e.,  that  (dA)  sin  <f>^pdpdd. 
Then  the  above  expression  can  be  written 

KPUr')csc4>fJ"p^dpde  =  f£-^'jr.  . 

Journal  Friction. — -In  general  it  is  not  known  how  the  nor- 
mal pressure  (and  hence  the  friction)  varies  over  the  surface  of 
a  journal.  It  is  customary  to  compute  the  "axle"  or  "journal 
friction ' '  from 

F  =  f'R, 

in  which  F  denotes  the  value  of  a  single  resistance  applied  to 
the  surface  of  the  journal  whose  moment  about  the  axis  is  the 
same  as  that  of  the  actual  frictional  resistance,  R  the  resultant 
pressure  between  journal  and  bearing,  and  /'  a  coefficient  of 
journal  friction. 

The  coefficient  /'  is  determined  from  f  =  F/R,  R  and  F  hav- 
ing been  experimentally  measured.  The  values  of  f  and  of  / 
(for  pivots)  depend  on  circumstances,  as  described  below.     They 


286  ROTATION.  [Chap.  XII. 

range  from  about  0.004  in  the  most  favorable  cases  to  about  0.08 
in  ordinary  lubrication. 

Friction  of  Lubricated  Surfaces."^ — "  The  laws  which  appear  to 
express  the  behavior  of  well-lubricated  surfaces  are  almost  the 
reverse  of  those  of  dry  surfaces."     Thus  the  frictional  resistance 

(i)  is  almost  independent  of  the  pressure  with  bath  lubrica- 
tion .  .  .  ; 

(2)  varies  directly  as  the  speed  for  low  pressures  .  .  .  ; 

(3)  depends  more  upon  the  temperature  than  on  any  other 

condition  .  .  .  ; 

(4)  with  flooded  bearings ,  depends  but  slightly  upon  the  nature 

of  the  material  of  which  the  surfaces  are  composed...; 

(5)  of  rest  is  enormously  greater  than  the  friction  of  motion... ; 

(6)  is  least  at  first,  and  rapidly  increases  with  the  time  after 

the  two  surfaces  are  brought  together.  .  .  . 

EXAMPLES. 

v/  I.  Show  that  the  value  of  the  frictional  moment  on  a  hollow 
flat  pivot  is  ^jN{r2^  —  r^^)/{r^  —  r^),  r^  and  fj  denoting  the  inner 
and  outer  radii  respectively  of  the  pivot. 

2.  Deduce  an  expression  for  the  frictional  moment  on  a 
pivot  formed  of  a  frustrum  of  a  cone,  there  being  no  pressure 
on.  the  lower  base  of  the  frustrum.  Use  notation  of  fig.  215(6) 
and  call  the  radius  of  the  step  r^  and  that  of  the  end  of  the 
pivot  fi. 

*  Abridged  and  quoted  from  Goodman's  "Mechanics  Applied  to  En- 
gineering." Chap.  VII  of  that  work  is  an  extensive  discussion  of  the 
subject  of  Friction. 


CHAPTER  XIII. 
ANY  PLANE  MOTION  OF  A  RIGID  BODY  (RESUMED). 

§  I.     General  Principles. 

279.  The  Effective  Forces. — Let  fig.  216  be  a  section  of  the 
moving  body  parallel  to  the  plane  of  the  motion,  and  P  and  P' 
any  two  points  of  the  body  in  that  section. 

Let  a'  denote  the  acceleration  of  P' ; 
a  the  angular  acceleration  of  the 

body ; 

(^  its  angular  velocity ; 

r  the  distance  P'P; 

dm    "   mass  of  the  particle  at  P. 

According  to  art.  220,  the  acceleration  of  P       ^^^  , 

can  be  resolved  into  three  components  as 

shown  in  the  figure,  one  being  the  same  as  the  acceleration  of 

P'  and  the  other  two  being  components  of  the  acceleration  of 

P  relative  to  P'.     Therefore  the  effective  force  for  the  particle 

at  P  can  be  resolved  into  three  components  whose  directions  are 

the  same  as  those  of  the  three  component  accelerations,  their 

values  being 

dm -a',     dm-ra,     and     dm-rw^. 

In  accordance  with  art.  221,  all  components  like  dm-a^  may  be 
called  the  translational  effective  forces  and  all  those  like  dm-ra 
and  dm-rco^  may  be  called  the  rotational  effective  forces. 

280.  Moment  of  the  Effective  System. — The  moment  axis  is 
taken  perpendicular  to  the  plane  of  the  motion  and  through  any 
point  of  the  body.  Let  P'  (fig.  217)  be  the  point  and  P'X  and 
P'Y  be  two  axes  parallel  to  the  plane  of  the  motion  and  fixed  in 
direction.  Now  the  moment  of  the  effective  force  for  the  par- 
,ticle  at  P  equals  the  sum  of  the  moments  of  its  components ;  the 

287 


288 


ANY  PLANE  MOTION  OF  A  RIGID  BODY,    [Chap.  XKI. 


moment  of  dm-roj^  is  zero,  that  of  dm-ra  is  {dm'ra)r^  and  that 
of  dm  •  a'  equals  the  sum  of  the  moments  of  its  x  and  y  compo- 
nents {dm-ax  and  dm -ay)  or  —{dm-ax)y  and  (dm'ay)x  re- 
spectively.    Hence  the  moment  of  the  effective  force  equals 

(dm  •  ra)r-\-{dm  •  ay)x  —  {dm  •  aj)y, 

and  the  sum  of  all  such  moments  (for  all  the  particles  of  the 
body)  is 

a  J  dm '  r^  +  ay' J  dm  •  x  —  dx^J  dm'y^ 


or 


la  +  may'x  —  max'y^ 


I  being  the  moment  of  inertia  of  the  body  with  respect  to  the 
moment  axis,  x  and  y  the  coordinates  of  the  mass-centre  at  the 
instant  for  which  the  moment  is  computed. 


Fig.  217. 

This  expression  for  moment  simplifies  considerably  for  two 
special  moment  axes,  as  follows: 

(a)  If  the  moment  axis  contains  the  mass-centre,  x  and  y  are 
zero,  and  the  expression  for  moment  reduces  to  la,  I  denoting 
the  moment  of  inertia  with  respect  to  that  axis. 

(6)  If  the  moment  axis  coincides  with  the  instantaneous  axis 
of  no  acceleration  (the  line  all  points  of  which  have  at  the  instant 
no  acceleration),  a/  and  ay  are  zero,  and  the  expression  for 
moment  reduces  to  la,  I  denoting  the  moment  of  inertia  of  the 
body  with  respect  to  that  axis. 

281.  Equations  of  Motion. — Let  the  motion  of  the  mass-cen- 
tre be  referred  to  a  set  of  fixed  axes,  x,  y,  and  z,  the  last  being 
perpendicular  to  the  plane  of  the  motion ;  also  let 


§1.]  GENERAL  PRINCIPLES,  28q 

a^  denote  the  x  acceleration  of  the  mass-centre ; 
ay  the  y  acceleration  of  the  mass-centre ; 
IFx    "    sum  of  the  x  components  of  the  external  forces; 
IFy  "       •*     **    •'  y  *'  "     "  "  " 

J:M    "      "     **    "       moments       "    "  "  "         about 

an  axis  through  the  mass-centre  and  perpendicular  to 
the  plane  of  the  motion. 
As  shown  in  art.  240 

i'F;c=Diaj,     I¥y=msLy]     also     2111=10:,     .     .     (i) 

since,  according  to  D'Alembert's  principle,  the  sums  of  the  mo- 
ments of  the  external  and  effective  forces  about  any  line  are 
equal. 

282.  Resultant  of  the  Effective  System  in  Important  Special 
Cases. — It  is  assumed  in  this  article  that  the  body  is  homogene- 
ous and  has  a  plane  of  symmetry  which  is  parallel  to  the  plane 
of  the  motion.  Imagine  the  body  divided  into  elementary  rods 
perpendicular  to  the  plane  of  the  motion.  As  explained  in  art. 
264,  the  effective  force  for  each  of  these  is  one  whose  action  line 
is  in  the  plane  of  symmetry;  hence  the  effective  forces  for  all 
the  rods  constitute  a  coplaner  system,  its  plane  being  the  plane 
of  symmetry. 

We  proceed  to  determine  the  resultants  of  the  effective  forces 
corresponding  to  translational  and  rotational  components  of 
the  motion.  In  general,  the  resultant  of  each  set  is  a  single 
force.  Let  fig.  218(a)  represent  the  section  of  symmetry  of  the 
body,  C  the  mass-centre  and  P'  any  other  point  of  the  section. 
In  addition  to  the  notation  of  the  foregoing  articles,  let 

r  denote  the  distance  from  C  to  P' ; 

a'  the  acceleration  of  C  relative  to  P',  i.e.,  its  acceleration  in  the 

rotational  component; 
k  the  radius    of    gyration   of  the   body   with    respect    to    the 

axis  through  P'  perpendicular  to  the  plane  of  the  motion. 

(a)  Regarding  the  motion  as  resolved  into  a  rotation  about 
the  axis  through  P'  and  the  corresponding  translation. — Accord- 
ing to  art.  244,  the  resultant  of  the  translational  effective  forces 
equals  ma'  and  acts  in  the  direction  of  a'  through  the  mass- 


:igO 


ANY ^ PLANE  MOTION  OF  A  RIGID  BODY,      [Chap.  XII. 


centre  as  shown.  According  to  art.  262,  the  resultant  of  the 
rotational  effective  forces  equals  ma'  and  acts  in  the  direction 
of  a'  through  a  point  Q  in  the  line'p'C  such  that  P'Q  equals 
k^/r.  This  last  force  may  be  resolved  into  two  components  nifa 
and  mrii?,  as  represented,  fig.  218(a). 

(6)  Regarding  the  motion  as  resolved  into  a  rotation  about 
the  axis  through  the  mass-centre  and  the  corresponding  trans- 
lation {P'  coincides  with  C). — The  resultant  of  the  translational 
effective  forces  is  a  force  equal  to  wid  and  acts  in  the  direction  of 
a  through  the  mass-centre  as  shown  in  fig.  218(6).  Accord- 
ing to  art.  262,  the  resultant  of  the  rotational  forces  is  a  couple 
whose  moment  is  /a. 


(a)  C6) 

Fig.  218. 

If  the  acceleration  of  the  mass-centre  and  the  angular  accel- 
eration of  the  body  are  both  zero,  the  resultant  of  the  effective 
forces  vanishes. 


§  II.     Applications. 

283.  Determination  of  the  Motion. — The  general  method 
consists  in  writing  the  equations  of  motion  for  the  case  in 
hand  and  deducing  from  thern  the  value  of  the  angular  accelera- 
tion of  the  body  and  that  of  the  acceleration  of  a  point  of  it. 
The  method  is  further  explained  in  the  solution  of  some  of  the 
following 

EXAMPLES. 

I.  A  homogeneous  cylinder  rolls  without  slipping  down  an 
inclined  plane.     Determine  the  motion. 


/ 


§  II.  ]  APPLICA  TIONS.  2  9 1 

Solution:    Let  W  denote  the  weight  of  the  cylinder,  and  m 
its  mass ;  also  let  F  and  A^  denote  the  com- 
ponents of  the  reaction  of  the  plane  along 
and  perpendicular  to  its  surface  (fig.  219).* 

The  external  forces  acting  on  the  cyl- 
inder are  F,  N,  and  W;  hence  with  coor- 
dinate axes  as  shown,  eqs.  (i),  art.  281  be- 
con^e  Fig.  219. 

IFx  ='-hW  sin  (j)—F  =  ma^,     IFy  =  N  —  W  cos  (})  =  mdy, 
IM  =  Fr  =  Ta. 

Evidently  dy  =  o,  hence  a  =  ax.  .  Now  v  =  rw  (see  ex.  2,  art.  220); 
hence  a  =  ra.  This  equation  and  the  first  and  third  above  deter- 
mine a"  and  a.     We  find  that  (since  I  =  ^mr^) 

a=^{g  sin  ^)/r, 
a=Jgsin0. 

We  may  also  determine  the  reaction  of  the  plane;  from  the 
first  equation 

F  =  ^W  sin  (j>     and  from  the  second,     N  =  W  cos  <^. 

^  2.  Show  that  there  will  be  slipping  between  the  cylinder 
and  plane  if  i  tan  (j)  is  greater  than  the  coefficient  of  static 
friction. 

3.  If  in  ex.  I  ^  =  30°  and  the  coefficients  of  static  and  kinetic 
friction  kre  respectively  0.25  and  0.2,  determine  the  angular 
acceleration  of  the  cylinder  and  the  linear  acceleration  of  its 
mass-centre. 

4.  A  homogeneous  cylinder  rests  on  a  smooth  horizontal 
plane,  and  a  horizontal  force  P  is  applied  by  means  of  a  string 
wrapped  about  it  (see  fig.  220).  Determine  the  angular 
acceleration  of  the  cylinder  and  the  acceleration  of  its  mass- 
centre. 

*  We  assume  in  this  and  the  following  examples  on  rolling  that  the 
rolling  body  and  the  surface  on  which  it  rolls  do  not  distort  each  other, 
thus  leaving  a  point  or  line  of  contact;  then  there  is  no  "rolling  resist- 
ance" (art.  285). 


292 


y4NY  PLANE  MOTION  OF  A  RIGID  BODY,    [Chap.  XIII. 


Solution :  The  external  forces  acting  on  the  cylinder  are  P, 
weight  {W),  and  the  resistance  {N)  of  the  surface.  Equations 
(i)  of  art.  281  become  (see  fig.  220),' 

IFx  =  P=nidxy     IFy  =  o  =  may,     IM  =  Pr  =  Ia, 


From  the  last  equation,  a=Pr/I,  and  from  the  first  ax  =  P/ni, 
Since  ay  =  o,  a  =  ax  =  P/m. 

^      5.  Solve  ex.  4,  supposing  that  the  plane  is  rough,  its  coeffi- 
cient of  kinetic  friction  being  /.      ^  '    "  ^ 

6.  Fig.  221  represents  a  wheel  rolling  on  a  horizontal  sur- 
face. If  the  distance  of  its  centre  of  gravity 
from  the  centre  is  c,  and  its  radius  of  gyration 
with  respect  to  an  axis  through  its  centre  and 
perpendicular  to  the  plane  of  the  motion  is  k, 
show  that 


z 

0 

^   X 

> 

rw 

p 

\          X 

^ 

•IZl 

Fig.  220. 


—gc  sin  ^  =  rcsin  ^•a»^  +  (r^  — 2rc  cos  d-\-k'^')a. 

284.  Kinetic  Reactions. — In  the  following 
examples  the  motion  of  the  body  under  consid- 
eration is  given  and  the  unknown  quantities  are 
forces .  These  are  determined  by  use  of  the  equa- 
tions of  motion,  but  sometimes  .other  forms  of  the  equations  are 
more  convenient.  These  other  forms  are  obtained,  like  those 
of  art.  281,  by  D'Alembert's  principle,  i.e.,  by 
expressing  algebraically  the  fact  that  the  ex- 
ternal forces  acting  on  a  body  are  equivalent 
to  the  resultant  effective  force  for  it.  This 
equivalence  can  be  expressed  in  many  ways; 
thus,  to  deduce  the  equations  of  art.  281,  we 
wrote  two  resolution  equations  and  one  mo- 
ment equation  with  moment-axis  through  the  mass-centre 
and  perpendicular  to  the  plane  of  the  motion.  But  we  may 
write  one  resolution  and  two  moment  equations  or  three  moment 
equations.  The  first  example  below  is  solved  from  two  sets  of 
equations,  thus  illustrating  different  forms  of  the  equations  of 
motion. 


IL] 


APPLICATIONS. 


293 


EXAMPLES. 

I.  A  wheel  whose  mass-centre  is  not  in  its  geometrical  axis 
is  rolled  along  a  horizontal  roadway  at  a  constant  speed  by  means 
of  a  horizontal  force  (P)  applied  at  its  centre.  Determine  the 
value  of  P  and  the  reaction  of  the  roadway  for  any  position  of 
the  wheel. 

Solutions:  (i)  By  use  of  the  equations  of  art.  281.  Let  fig. 
222  represent  the  wheel,  P'  being  its  centre  and  C  its  mass-cen- 
tre, and  let  F  and  N  denote  the  hori- 
zontal and  vertical  components  of  the 
reaction  of  the  roadway.  The  exter- 
nal forces  acting  on  the  wheel  are  W, 
P,  F,  and  N;  hence  eqs.  (i),  art.  281, 
become 


IF,  =  P-F 


IFy=N 


max, 
W  =  ma, 


(I) 
(2) 

(3) 


IM  =  F{r+c  sin  d)  -Nc  cos  d 
—  Pc  sin  ^  =  0.  .  '.  . 
Now  the  acceleration  of  C  (a)  equals  the  vector  sum  of  the  accel- 
eration of  P'  and  that  of  C  relative  to  P'  (art.  196).  Since  the 
wheel  rolls  uniformly,  the  acceleration  of  P'  is  zero  and  that 
of  C  relative  to  P'  is  coj^  in  the  direction  CP' ;  hence 

a  =  C(i?  directed  along  CP' . 
Also  a^;  =  co;^  cos  <9     and     a^  =  —  co;^  sin  ^, 

and  eqs.  (i),  (2),  and  (3)  may  be  written 

F  =  P  —  mcco^  co?,d,     N^W  —  mcio^sXnd,    and 
{P  —  mcco'^co^d){r-\-cs,m.d)  —  {W  —  mcco'^  sin  d)c  cos  O—Pc  s,m.d  —  o. 
From  these  we  find  that 


and 


P  =  {Wc/r-\-mcco'^)  cos  ^, 
F  =  W{c/r)  cos  d, 
N ^W  —  mcoj^  sin  6. 


(2)  The  three  components  of  the  resultant  effective  force  as 
described  in  art.  282,  case  (a),  are  wa',  mra,  and  mrto^.     P' 


294  ANY  PLANE  MOTION  OF  A  RIGID  BODY,    [Chap.  XIIL 

(fig.  2 1 8a)  being  chosen  at  the  centre  of  the  wheel,  a'  is  zero; 
and  since  a  is  zero,  the  resultant  effective  force  is  inraj^  =  mcoj^ 
directed  from  C  to  P'.  Now  by  D'Alembert's  principle  F,  A^, 
P,  and  W  are  equivalent  to  mcco^,  or  they  are  in  equilibrium  with 
mcco^  reversed.     Hence 

IMa  =Pr  —  Wc  cos  6  —  mcco^r  cos  ^  =  o, 
IMp,  =  Fr-Wc  cos  d=^o, 
IFy  =  N -W-\-  mco)''  sin  d  =  o. 

From  these  (without  elimination)  we  get  the  same  values  for  P, 
Fj  and  N  as  were  found  in  ( i ) . 

The  kinetic  reaction  (on  the  wheel)  is  vertical  and  equals 
mcoj^  sin  d ;  it  acts  downwards  for  values  of  6  between  o°  and 
i8o°  {C  is  above  P')  and  upwards  for  values  of  6  between  i8o° 
and  360°  {C  is  below  P')-  Otherwise  stated,  when  C  is  above 
P',  N  is  less  than  W,  and  when  C  is  below  P',  N  is  greater  than 
W.  The  kinetic  component  of  the  reaction  A^  is  called  "  ham- 
mer blow"  in  locomotive  parlance. 

2.  Show  that  the  wheel  of  ex.  i  lifts  from  the  roadway  in  a 
certain  position  if  the  speed  of  the  centre  of  the  wheel  is  greater 
than  rVg/c. 

3.  In  which  position  of  C  is  A/"  a  maximum,  and  what  is  that 
value  if  the  velocity  is  slightly  less  than  r\^ g/c. 

4.  A  steam-engine  connecting-rod  with  no  piston-rod  attached 
is  drawn  by  the  crank.  It  is  required  to  determine  the  kinetic 
reactions  at  the  ends  of  the  rod  in  any  position. 

Solution:  Let  AB  and  P'B  (fig.  223)  be  the  crank  and  rod 
respectively,  C  being  the  mass-centre  of  the  latter.  Let  r  de- 
note the  length  CP',  a  the  acceleration  of  P',  m  the  mass  of  the 
rod,  a  and  co  its  angular  acceleration  and  velocity  respectively, 
and  k  the  radius  of  gyration  of  "the  rod  with  respect  to  a  line  per- 
pendicular to  the  plane  of  the  motion  at  P'. 

We  first  determine  the  resultant  effective  force  {R)  for  the 
rod.     Its  three  components  as  given  by  art.  282  are 

ma,     nira,     and     mroj^, 
acting  as  shown  in  the  figure,  the  distance  of  Q  from  P'  bein;^ 


§  II.]  APPLICATIONS.  295 

k'^/r.     Supposing  these  to  have  been  computed,  R  can  be  readily 
found  graphically  (the  construction  is  not  shown). 

Let  V  and  P  denote  the  reactions  on  the  rod  at  P'  and  B 
respectively;    the  "guide-rods"  being  supposed  frictionless,   V 


Fig.  223. 

is  vertical.  Since  F,  P,  and  R  reversed  are  in  equilibrium,  their 
action  lines  intersect  in  a  point.  So  produce  R  to  intersect  V 
and  join  that  intersection  with  B\  this  line  is  the  action  line 
of  P.  To  determine  the  values  of  V  and  P,  we  draw  a  force 
triangle  for  V,  P,  and  R  reversed.  Thus,  lay  off  MN  to  repre- 
sent the  magnitude  of  R,  and  from  N  draw  a  vertical  line  to  inter- 
sect the  action  line  of  P,  marking  that  point  O;  NO  and  OM 
respectively  represent  the  magnitudes  and  directions  of  V  and  P. 

5.  Let  r  and  c  denote  the  lengths  of  the  connecting-rod  and 
crank  respectively,  and  co^  the  angular  velocity  of  the  crank 
assumed  constant.  Take  r/c  =  4,  F  =  r/2 ,  ^^  =  r"^/^ ,  and  compute 
the  values  of  V  and  P  when  6  (see  fig.  223)  equals  0°,  30°, 
60°,  90°,  120°,  150°,  and  180°.  Also  plot  the  values  of  V  at  the 
corresponding  positions  of  P'  and  represent  each  P,  scaling  it 
from  the  corresponding  position  of  B\  then  draw  smooth 
curves,  joining  the  ends  of  the  vectors  representing  V  and  P 
thus  drawn.  I 

(For  a  method  of  computing  a,  a,  and  a),  see  art.  181  and 
ex.  I,  art.  219.  Notice  that  0  of  fig.  154  ^equals  ^  —  90°  of  figs. 
175  and  223.) 

285.  Rolling  Resistance. — Let  fig.  224  represent  a  wheel  or 
cylinder  rolling  upon  a  horizontal  surface  or  roadway  at  con- 
stant speed.     Let  W  denote  its  weight,  P,  applied  as  shown, 


296  ANY  PLANE  MOTION  OF  A  RIGID  BODY.    [Chap.  XIII. 

the   force  required  to   maintain   the  constant    speed,   and  R 

the  resultant  reaction   of  the  roadway.      Since   the   accelera- 

^ ...^^^         tion  of  the  mass-cen^e  and  the  angular  accel- 

/  \     eration  of  the  wheel  are  zero,  the  resultant 

/  w  \ 

/  \  effective  force  for  the  wheel  is  zero  (art.  282); 

L___r ^  ^ 1  hence  the  external  forces  W,  P,  and  R  are  in 

pi 

/  equilibrium,  and  R  must  act  through  the  cen- 

/  tre  and  be  inclined  as  shown. 

^^  The  horizontal  component  of  the  reaction 

of  the  roadway  is  called  "  rolling  resistance," 

_^"''"~  also  "  rolling  friction."    The  three  forces  being 

■  ^^^'  in  equilibrium,  the  horizontal  component  of  R 

equals  P,  and  hence  an  expression  for  P  is  also  a  value  of  the 

rolling  resistance.     If  a  denotes  the  distance  from  the  vertical 

through  the  centre  to  the  intersection  of  the  action  line  of  R  and 

the  circumference  of  the  wheel,  and  if  a  is  small  (as  it  is  except 

on  soft  roadways),  then  approximately 

Pr  =  Wa,     or     P  =  {a/r)W. 

From  this  equation  the  rolling  resistance  in  a  given  case  can  be 
computed  if  a  is  known. 

Like  the  coefficients  of  friction,  a  is  determined  from  exper- 
iment. Thus,  suppose  that  the  force  P  for  a  given  wheel  and 
roadway  has  been  measured;  that  value  and  the  weight  and 
radius  of  the  wheel  substituted  in  either  preceding  equation 
determine  a  for  the  case  in  hand.  The  distance  a  is  sometimes 
called  the  "coefficient  of  rolling  resistance."  Observe  that  it  is 
not  an  abstract  number  like  the  coefficients  of  friction,  but  a 
length. 

Practically  no  general  facts  or  laws  are  known  concerning 
the  coefficient  of  rolling  resistance.  The  coefficient  is  usually 
regarded  as  independent  of  the  weight  (if  moderate  so  that  no 
permanent  deformation  of  roller  or  roadway  occurs),  but  there 
is  disagreement  as  to  the  relation  between  it  and  the  radius — 
it  being  held,  for  example,  that  a  is  independent  of  r,  also  that 
a  varies  as  V^. 

The  following  values  are  given  to  afford  a  notion  of  the  values 
of  a  in  a  few  cases. 


§11 J                                           APPLICATIONS.                                              297  I 

Rollers  of  elm  on  an  oak  track  (Coulomb) 0.032  in.  \ 

Iron  or  steel  wheels  on  iron  or  steel  rails 0.007-0.020    "  J 

"     •'     "          "       "wood 0.06-0.10     "  \ 

\ 

EXAMPLE. 

If  3everal  rollers  are  used  for  rolling  a  heavy  body  along  a  \ 

horizontal  roadway,  show  that  their  rolling  resistance  is  given  \ 

by  W{a'  +  a")/2r,  W  denoting  the  weight  of  the  body,  r  the  \ 

radii  of  the  rollers,  and  a'  and  a"  their  "coefficients  of  rolling  1 

resistance"  for  the  surfaces  at  which  the  rolling  occurs.  I 


CHAPTER  XIV. 
WORK    AND    ENERGY. 

§  I.     Work. 

286.  Work  Defined. — Work  is  said  to  be  done  upon  a  body 
by  a  force  when  the  application  point  is  displaced  so  that  the 
displacement  has  a  component  along  the  action  line  of  the  force. 

If  the  force  is  constant  in  direction,  the  projection  of  the 
displacement  on  the  action  line  of  the  force  is  called  effective 
displacement.  If  the  force  changes  its  direction  during  a  displace- 
ment of  its  application  point,  then  the  projection  of  the  dis- 
placement which  occurs  during  an  element  of  time  on  the 
corresponding  action  line  of  the  force  is  called  the  effective 
displacement  for  that  elementary  interval. 

287.  Expressions  for  Work  Done  by  a  Force. — I.  The  force  is 
constant  in  magnitude  and  direction.  The 
amount  of  work  done  is  measured  by  the 
product  of  the  force  and  the  effective  dis- 
placement. Thus,  if  AB  (fig.  225)  is  a  dis- 
placement of  the  application  point  of  a  force 
F,  and  if  cj)  denotes  the  angle  between  AB 
and  F,  and  w  the  work  done  by  F,  then 

Fig.  225.  w  =  F(AB  cos  cl))  =  (F  cos  <j))AB.  .     (i) 

li  <j)  =  o,w  =  F  AB]  a  (j)  =  go°,w  =  o. 

Observe  that  in  the  last  form  of  ( i ) ,  the  expression  for  work 
is  the  product  of  the  component  of  the  force  along  the  displace- 
ment and  the  displacement. 

II.  The  force  varies  in  magnitude  or  direction  or  both.  Let 
AB  (fig.  226)  be  a  portion  of  the  path  of  the  application  point  of 
the  force  F,  P  being  any  intermediate  position,  <j)  the  angle  be- 
tween F  and  the  tangent  to  the  path  at  P,  and  5  the  distance 
of  P  from  some  fixed  origin  in  the  path,  it  being  measured  posi- 

298 


§1.]  fVORK.  299 

lively  in  the  direction  of  the  motion.     Then  if  dw  denotes  the 
work  done  by  the  force  while  its  applica- 
tion point  describes  an  elementary  portion 
of  the  path  ds  including  P, 

dw  =  F  ds  cos  (j) ; 

and  if  w  denotes  the  work  done  by  F  dur- 
ing the  displacement  AB  in  which  s 
changes  from  s'  to  5", 


w 


=  r"  F  ds  cos  (l>  =  £" Ftds (2) 


li  <j)  =  o,Ft  =  F\  a  (j)  =  go°,  Ft  =  0  and  w  =  o. 

288.  Sign  of  a  Work. — It  is  convenient  to  give  sign  to  the 
work  done  by  a  force.  The  rule  is  as  follows:  A  work  is  re- 
garded as  positive  or  negative  according  as  the  effective  dis- 
placement agrees  with  or  is  opposed  to  the  force  in  sense.  It 
must  be  remembered  that  a  displacement,  and  hence  its  pro- 
jection also,  is  a  vector  quantity. 

If  the  angle  (j)  is  always  taken  as  in  figs.  225  and  226,  i.e.,  be- 
tween the  portions  of  the  lines  representing  the  force  and  the 
displacement  toward  which  the  arrows  point,  then  the  expres- 
sions for  w  in  the  preceding  article  give  the  correct  sign  of  the 
work. 

289.  Unit  of  Work. — Equations  (i)  and  (2),  art.  287,  imply 
as  unit  the  work  done  by  a  unit  iorce  "  acting  through  "  unit  dis- 
tance. The  value  of  the  unit  hence  depends  on  the  units  used 
for  force  and  distance.  Thus  we  have,  corresponding  to  the 
pound  and  foot,  the  foot-pound  unit  of  work,  to  the  kilogram 
and  meter,  the  meter-kilogram,  to  the  dyne  and  centimeter,  the 
dyne-centimeter,  etc.  The  last-named  unit  has  also  a  special 
name,  erg* 

290.  Work  Diagram.— If  values  of  Fi  and  5  be  plotted,  on  tv/o 
rectangular  axes  (see  fig.  227)  for  all  positions  of  the  application 
point  of  the  force  F,  the  curve  joining  the  plotted  points  might 
be  called  a  "tangential  force-space  (or  FfS)  curve."     The  por- 

*  For  dimensions  of  a  unit  work,  see  Appendix  C 


300  "     IVORK  AND  ENERGY.  [Chap.  XIV. 

tion  of  the  figure  between  the  curve,  the  5  axis,  and  any  two 
ordinates  is  called  a  work  diagram. 

Proposition. — The  area  of  a  work  diagram  represents  the  work 
done  by  the  force  during  the  displace- 
ment corresponding  to  the  bounding 
ordinates. 

Proof:  According  to  eq.  (2),  art.  287, 

^  Ftds.    Since  Ff  and  5  also  denote 

coordinates   of  points  on  the  Ft-s  curve 
Fig.  227.  (fig.  227),  the  area  of  the  work  diagram 

IS    /     Ftdsy  as  is  shown  in  works  on  calculus.     Hence  the  area 

(according  to  some  scale)  equals  w.  Obviously  the  scale  accord- 
ing to  which  the  area  of  a  work  diagram  is  to  be  interpreted 
depends  on  the  scales  used  in  representing  Ft  and  s.  Thus  if 
one-inch  ordinates  and  abscissas  represent  100  lbs.  and  10  ft., 
respectively,  one  square  inch  of  area  represents  1000  ft. -lbs.  of 
work. 

Since  the  area  of  a  work  diagram  equals  the  product  of  the 
average  ordinate  and  the  base,  the  work  done  by  a  force  equals 
the  average  value  of  the  tangential  component  and  the  length 
of  the  path  described  by  the  application  point.  If  the  Ft-s  curve 
is  straight,  the  average  value  of  Ft  is  the  mean  of  the  initial  and 
final  values  and  the  computation  of  the  work  is  simple. 

EXAMPLES. 

I.  Fig..  228  represents  a  body  on  a  horizontal  surface  to  which 
two  forces  (P  and  Q)  are  applied  as  shown.     Compute  the  work 

— •>,    ^ 
w 

p= 100  lbs. 


a! °:^rhg 


//MW/M/M/W//'^^^^^^^^^^ 


Fig.  228. 


5=0         ^     r  s=5ffi^ 


done  by  these  forces,  the  weight,  and  the  resistance  of  the 
surface  while  the  body  moves  from  A  to  S  (5  ft.),  Q  being  a  vari- 
able force  so  that  Q  (in  lbs.)  =  25  (in  ft.)  and  the  friction  30  lbs. 
Solution:  Since  the  displacements  of  the  application  points 


§1.] 


IVORK. 


301 


of  W  and  A^  are  normal  to  the  action  lines  of  the  forces,  W  and 
N  do  no  work.  Since  F  and  P  are  constant  in  magnitude  and 
direction,  we  use  eq.  (i)  art.  287  for  computing  the  work  done  by 
them;  for  P,  cj)  is  zero  and  for  F,  <j>  is  180°,  hence 

the  work  done  by  P  =  100 -5  cos  0  =  500  ft. -lbs.,  and 
"       "         "      "   F=  30-5  cos  180°== —150  ft. -lbs. 
Since  Q  varies  in  magnitude  we  use  eq.  (2)  art.  287,  and  0 
being  180°,  the  expression  for  work  done  b}''  Q  is 

J]  Q  cos  180°  ds=  -  r  25  ds=-2$  ft.-lbs. 

This  value  can  be  readily  computed  from  the  average  value  of 
Q  which  is,  since  Q  varies  uniformly  with  s,  the  mean  of  ite 
initial  and  final  values;  these  are  o  and  10  lbs.  respectively » 
Hence  the  average  value  of  Q  is  5  lbs.,  and  as  it  acts  through 
5  ft.,  the  amount  of  work  done  by  Q  equals  5X5  =  25  ft. -lbs. 
The  sign  of  the  work  is  negative  because  the  senses  of  the  force 
and  displacemxent  are  opposite. 

2.  Fig.  229  represents  a  body  upon  an  inclined  plane  to 
which  two  forces  (P  and  Q)  are  applied  as  shown.  Compute 
the  works  done  by  them,  the  weight,  and  the  reaction  o£  the 
plane  while  the  body  moves  from  A  to  B  (10  ft.),  the  frictional 
resistance  of  the  plane  and  the  weight  being  10  and  100  lbs. 
respectively.  Ans.  Total  work  done=  —26.8  ft.-lbs. 


Fig.  229. 


Scales:  Horizontal,  1=^';  Vertical,  1=  600,000  lbs. 

Fig.  230. 


3.  In  punching  a  hole  (2  in.  diameter)  in  a  certain  iron  plate 
(ij  in.  thick),  the  pressure  (P)  between  punch  and  plate  varied 
as  the  ordinates  to  the  curve  of  fig.  230  (the  initial  values  of  P 
are  at  the  left).     Estimate  the  area  of  the  work  diagram  and 


30  2  ^.     IVORK  y4ND  ENERGY.  [Chap.  XIV. 

the  amount  of  work  done  by  the  punch  on  the  plate  in  the  oper- 
ation. Also  compute  the  work  from  the  average  value  of  the 
force.  » 

^/'  4.  One  end  of  an  elastic  cord  whose  natural  length  is  10  ft. 
is  fastened  to  a  body  on  a  horizontal  surface  and  the  other  end 
to  a  fixed  point  in  the  surface  20  ft.  from  the  body.  The  ten- 
sion in  the  cord  is  observed  to  be  30  lbs.  When  the  body  is 
released  it  moves  toward  the  fixed  point.  Draw  the  work  dia- 
gram for  the  tension  in  the  cord  and  determine  how  much  work 
is  done  by  the  tension  in  the  first  and  second  5  feet  of  the  dis- 
placement. 

N  5.  Suppose  that  a  gas  expands  behind  a  piston  in  a  cylinder 
according  to  the  law  pv^C,  C  being  a  constant,  p  the  gas  pres- 
sure per  unit  area,  and  v  the  volume  of  the  gas.  Show  that. the 
work  done  by  the  gas  on  the  piston  in  an  expansion  from  a 
volume  v^  to  a  volume  v^  equals  C  ^og£{vJv^. 
V  6.  Show  that  the  work  of  a  central  force  (one  always  directed 
toward  a  fixed  point)  in  any  displacement  of  its  application 

point  equals  —  j^Pdr,  in  which  P  denotes  the  general  value  of 

the  force,  i.e.,  its  value  when  its  application  point  is  any  dis- 
tance r  from  the  fixed  point  and  r^  and  rj  denote  the  values 
of  r  at  the  beginning  and  end  of  the  displacement  respect- 
ively. 

Solution:  Let  C,  fig.  231,  be  the  fixed  point  toward  which  P 

acts  and  OAB  the  path  of  the 
application  point  of  P.  The 
value  of  the  work  done  by  P  is 
given  by  eq.  (2)  art.  287.  Since 
(see  the  figure)  dr=  —ds  cos  <^, 
the  value  of  the  work  as  given 
by*e(i.  (2)  reduces  to  that 
given  above. 

V  7.  Suppose  that  the  body 
described  in  ex.  4  is  moved  (after 
the  cord  is  attached  to  it  as  described)  so  that  the  point  of  ap- 
plication of  the  cord  moves  in  the  circumference  of  a  circle 
whose  diameter  is  the  cord  in  its  first  position.     Compute  the 


§1.]  IVORK.  303 

work  done  by  the  tension  while  the  application  point  describes 
the  first  quarter  circle. 

u  8.  How  much  work  is  done  by  the  cord  up  to  the  instant 
when  it  resumes  its  natural  length? 

291.  Work  Done  by  Gravity  Upon  a  Body  in  Any  Motion. — 
Proposition. — The  work  done  by  gravity  upon  a  body  in  an^^ 
motion  equals  the  product  of  its  weight  and  the  vertical  distance 
described  by  the  centre  of  gravity,  and  the  work  is  positive  or 
negative  according  as  the  centre  of  gravity  has  descended  or 
ascended. 

Proof:  Let  Wj,  w^,  etc.,  denote  the  weights  of  the  particles 
of  the  body,  ;y/,  y^  y  etc.,  their  distances  ^^,^" 

above  some  datum  plane  (below  which        „/  ^^^^     ' 

the  body  does  not  descend)  at  the  begin-  }    l\       /  ^,4, 

ning  of  the  motion,  and  y/',  y^'',  etc.,        y'l  ^15'/    |" 
their  distances  above  that  plane  at  the         1     y']       j^i" 
end  of  the  motion  (see  fig.  232  where 


1/; 


a' a''  is  the  path  of  the  first  particle,  ^^^'  ^32- 

b'b'^  that  of  the  second,  etc.).  Also  let  W  denote  the  weight 
of  the  body  and  y  and  y''  the  initial  and  final  heights  of  its 
centre  of  gravity  above  the  plane.  Then  the  sum  of  the  works 
done  by  gravity  on  all  the  particles  is 

'i^iiyi  -yi')+'i^2(y2  -y2')+  •  •  •  =i'^'iyi+'^2y2+  •  •  •) 

-(i£/i:v/'+w2:v/'+  . . .) 

According  to  art.  64, 
w^y/'  +iV2y2'  +  ■  .  .  =Wy''     and     (w^yi  +W2y2'  +  .  .  .  )  =  Wy; 
hence  the  sum  of  the  works  done  on  all  the  particles  equals 

Wy'-Wy''  =  W(y-y'').  q.e.d- 

292.  Work  Done  by  Concurrent  Forces  and  by  Their  Result- 
ant.— Proposition. — The  work  done  by  any  number  of  concur- 
rent forces  in  a  displacement  of  their  application  point  equals 
that  done  by  their  resultant  in  that  displacement.* 

Proof:  Let  F^,  F^,  etc.,  denote  the  forces,  R  their  resultant, 
and  ^1,  (f)2,  etc.,  and  ^  respectively  the  angles  which  the  forces 

*  It  is  assumed  that  the  forces  and  their  resultant  have  a  common 
application  point. 


304- 


IVORK  AND  ENERGY. 


[Chap.  XIV. 


and  their  resultant  make  with  the  tangent  to  the  path  of  the 
application  point  (taken  as  explained  in  art.  288).  Then  ac- 
cording to  art.  36, 

R  cos  (j>  =  F^  cos  01+^2  cos  9^2+  •  •  •  • 
Therefore,  R  cos  (j)'ds  =  F^  cos  ^^ •  J5  +  F2  cos  <f)2-ds-\-  .  .  .  , 
and  / R  cos  ^'ds  =  J  F^  cos  (f)^'ds+  j  F^  cqs  9S2 •  <i5  +  ...  ; 

hence,  etc. 

293.  Work  Done  by  a  Pair  of  Equal,  Opposite,  and  CoUinear 
Forces. — Suppose  that  A  and  B  (fig.  233)  are  the  application 


Fig.  233. 

points  of  the  forces  at  any  instant  during  the  motion  and  let  P 
denote  the  value  of  the  forces  then.  Also  let  x'  and  y'  denote 
the  coordinates  of  A  and  ^"  and  y^'  those  of  B,  and  suppose  that 
A  moves  from  A  ^  to  A  2  and  B  from  B^to  B^/^.Q.j  assume  for  sim- 
plicity that  the  displacements  are  coplanar.  The  discussion  can 
be  easily  extended  to  include  non-coplanar  displacements. 

The  work  done  by  each  force  equals  the  sum  of  the  works 
done  by  its  x  and  y  components;  for  the  force  P  acting  on  A 
these  are 

-  fp  cos  d'dxr   and     -fPsmd-dy^ 

and  for  the  force  P  acting  on  B  they  are 

JPcosd'dx"     and     fPsind-dy. 

The  work  done  by  both  forces  equals 

/P[cos  d . (dx^'-dx')  +sin  d-idy^-d/M 

It  is  plain  from  the  figure  that 

r2=(:^"-A;')'+(y'-y)^ 


§1.]  IVORK,  iC5 

hence       r  dr^{x''  -x'){dx''  -dx')-V{y'  -y){dy'  -dy), 

or  dr  =  cos6'  {dx"  -dx')+  sin  6  -  {dy  -  dy) . 

Substituting  according  to  this  relation  in  the  expression  for 
total  work  we  find  that  the  latter  becomes 

iP  dr  (when  the  force  P  on  A  acts  from  B  to  .4),  but 

Jri 

—  /  ^P  dr  (when  the  force  P  on  ^  acts  from  A  to  B), 

Jrx 

as  will  be  seen  by  changing  the  arrows  on  P  in  the  figure  and 
making  the  necessary  changes  in  the  discussion. 

If  the  distance  between  the  application  points  of  the  forces 
does  not  change  during  the  displacement,  dr  =  o,  and  the  work 
done  by  the  pair  of  forces  equals  zero.  If  P  depends  only  on  r, 
then  the  work  done  by  P  depends  only  on  the  initial  and  final 
values  of  r  and  not  at  all  on  the  way  in  which  r  changes  during 
the  displacement. 

EXAMPLE. 

How  much  work  is  done  by  the  steam  in  one  cylinder  of  a 
locomotive  during  one  stroke  of  the  piston  ? 

Solution :  Consider  a  forward  stroke  of  the  piston  and  let  P 
denote  the  pressure  on  the  piston  when  its  distance  from  the 
rear  end  of  the  cylinder  is  r.  Then  the  work  done  by  the  pres- 
sures on  the  piston  and  rear  end  of  the  cylinder  in  one  stroke 
equals 

rPdr  =  Pa(r,-r,)  =  P,s, 

Tj  and  rg  denoting  the  values  of  r  at  the  beginning  and  end  of  the 
stroke,  Pa  the  average  value  of  the  steam  pressure,  and  ^  the 
length  of  stroke. 

This  value  of  the  work  might  also  be  obtained  by  computing 
the  work  done  by  each  pressure  separately.  Thus  let  R  denote 
the  radius  of  the  driving-wheels,  then  the  distance  through  which 
the  locomotive  moves  in  one  stroke  equals  tiR,  and  supposing 
that  the  locomotive  is  running  forward,  the  work  done  by  the 
steam  on  the  rear  end  of  the  cylinder  equals  —PaTrR,  and  that 
done  on  the  piston  equals  PairtR+s);  hence  the  work  done  by 
both  pressures  equals  Pa  s. 


3o6  IVORK  AND  ENERGY.  [Chap.  XIV. 

294.  Work  Done  by  a  Body  Against  a  Force. — It  is  con 

venient  to  use  the  expression  "work, done  by  a  body  against  a 
force"  applied  to  it;  we  mean  by  it  the  negative  of  the  work 
done  upon  the  body  by  the  force.  Thus  if  a  body  weighing  10 
lbs.  is  made  to  rise  5  ft.,  gravity  does  —50  ft. -lbs.  of  work  upor 
it  and  the  body  does  +50  ft. -lbs.  of  work  against  gravity. 

In  accordance  with  the  above,  when  the  sense  of  a  force  and 
that  of  effective  displacement  of  its  application  point  are  oppo- 
site, the  work  done  by  the  body  is  positive ;  when  they  are  the 
same,  the  work  done  by  the  body  is  negative. 

§  11.     Energy. 

295.  Energy  Defined. — When  the  state  or  condition  of  a  body 
is  such  that  it  can  do  work  against  forces  applied  to  it,  the  body 
is  said  to  possess  energy.  A  stretched  spring  can  do  work  against 
forces  applied  to  it  if  they  are  such  that  it  may  contract ;  a  body 
in  motion  can  do  work  against  an  applied  force  which  tends  to 
stop  it.     The  spring  and  the  body  therefore  possess  energy. 

The  amount  of  energy  possessed  by  a  body  at  any  instant  is 
the  amount  of  work  which  it  could  do  against  applied  forces 
while  its  state  or  condition  changes  from  that  of  the  instant  to 
an  assumed  standard  state  or  condition.  The  meaning  of  the 
standard  condition  is  explained  in  subsequent  articles. 

The  unit  of  energy  must,  in  accordance  with  the  above,  be 
the  same  as  the  unit  of  work. 

296.  Kinetic  Energy  Defined.  —  Energy  is  classified  into 
kinds  depending  on  the  state  or  condition  of  the  body  in  virtue 
of  which  it  has  energy.  Kinetic  energy  of  a  body  is  energy 
which  it  has  by  virtue  of  its  velocity. 

297.  Kinetic  Energy  of  a  Particle. — The  amount  of  kinetic 
energy  possessed  by  a  particle  at  any  instant  is  the  work  which 
it  could  do  while  the  velocity  changes  from  its  value  at  that  in- 
stant to  some  other  value  taken  as  a  standard.  It  is  customary 
to  take  zero  velocity  as  the  standard  one;  this  being  understood, 
we  may  say  that  the  amount  of  kinetic  energy  possessed  by  a 
particle  is  the  work  which  it  can  do  in  "  giving  up  its  velocity." 

Proposition. — The  kinetic  energy  of  a  particle  whose  mass 
and  velocity  are  m  and  v  respectively  equals  ^Jiv^, 


§  II.]  ENERGY.  307 

Proof :  Let  R  denote  the  resultant  of  all  the  forces  acting  on 
the  particle  while  it  "gives  up  its  velocity."  Let  A  and  B  de- 
note the  beginning  and  end  of  the  path,  and  s'  its  length.  Then, 
according  to  arts.  292  and  287,  the  Vork  {w)  done  by  all  the 
forces  on  the  particle  in  the  motion  from  ^  to  B  is  given  by 


w 


=/>■■* 


Hence  the  work  done  'oj  ^he  particle  against  the  forces,  or  the 
kinetic  energy  {E)  of  the  particle,  is  given  by 

E=-f'Rtds, 
Now  Rt  =  mat  =  m  dv/dt    (see  art.  236) ; 

hence  E=  —  jmv  dv  =  ^mv*,  Q.B.D. 

298.  Kinetic  Energy  of  any  System  of  Particles. — The  kinetic 
energy  of  a  system  of  particles  equals  the  sum  of  the  kinetic 
energies  of  the  separate  particles.  If  m  and  v  denote  the  mass 
and  velocity  respectively  of  any  particle  of  a  s)cstem,  and  E  the 
kinetic  energy  of  the  system, 

E  =  ii'mv» •    .     (i) 

I.  Translating  Body. — In  this  case  all  particles  have  at  each 
instant  equal  velocities,  hence 

or,  if  M  denote  the  mass  of  the  body,  the  kinetic  energy  E  is 
given  by 

E=iMv». (2) 

II.  Rotating  Body. — In  this  case  the  velocity  of  any  particle 
of  the  body  equals  the  product  of  its  distance  from  the  axis  of 
rotation  and  the  angular  velocity  of  the  body  (art.  216).  Let  r 
denote  the  distance  of  any  particle  from  the  axis  and  cu  the 
angular  velocity  of  the  body;  then  the  vaVrte"of  the  kinetic 
energy  is 

Now  Imr^  is  the  moment  of  inertia  of  the  rotating  body  with 


3^^  IVORK  AND  ENERGY.  [Chap.  XIV. 

respect  to  the  axis  of  rotation;  if  /  be  used  to  denote  this  quan- 
tity, the  kinetic  energy  is  given  by        ♦, 

.E  =  iW (3) 

III.  Body  having  any  Plane  Motion. — In  this  case  the  state 
of  the  motion  at  any  instant  may  be  regarded  as  rotational  (see 
arts.  2  22  and  223).  As  shown  in  art.  223,  the  velocity  of  any 
particle  of  the  body  at  any  instant  equals  the  product  of  its  dis- 
tance from  the  line  which  is  the  axis  of  rotation  at  that  instant 
(instantaneous  axis)  and  the  angular  velocity  of  the  body.  The 
reasoning  in  Case  II  applies  here  if  the  word  **  instantaneous  "  is 
inserted  before  the  word  "  axis  " ;  then  ^loj^  is  the  expression  for 
the  kinetic  energy  of  a  body  having  any  plane  motion,  /  being 
the  moment  of  inertia  with  respect  to  the  instantaneous  axis. 

Since  the  instantaneous  axis  in  general  moves  about  in  the 
moving  body,  the  expression  above  is  not  always  convenient  to 
apply,  and  another,  though  not  so  simple  in  form,  is  simpler  in 
its  application.  This  expression  may  be  deduced  as  follows: 
In  addition  to  the  notation  employed  in  the  preceding,  let  / 
denote  the  moment  of  inertia  of  the  body  with  respect  to  a  cen- 
tral axis  perpendicular  to  the  plane  of  the  motion,  v  the  velocity 
of  the  mass-centre  at  the  instant  considered,  r  the  distance  be- 
tween the  mass-centre  and  the  instantaneous  axis,  and  M  the 
mass  of  the  body.     Then 

I  =  T-i-AIr^  (art.  256)     and    v^roj; 
hence                ^Ioj^^^Foj^'  +  ^MT^w^ 
or  E  =  iIa>2  +  ^Mv2 (4) 

Now  ^Ico^  is  the  kinetic  energy  which  the  body  would  have  if 
rotating  about  a  fixed  axis  through  its  mass-centre  with  an  angu- 
lar velocity  oj,  and  ^Mv^  is  the  kinetic  energy  which  it  would 
have  if  translating  with  a  velocity  v.  Hence  the  kinetic  energy 
of  a  body  having  any  plane  motion  is  regarded  as  consisting 
of  two  parts,  and  they  are  called  rotational  and  translational. 

EXAMPLES. 

I.  Express  the  kinetic  energy  of  a  body  weighing  1.5  tons 
and  moving  at  a  speed  of  60  mi.-per-hr.  in  ft.-lbs. 


§11.]  ENERGY.  309 

2.  Express  the  kinetic  energy  of  a  cylindrical  disk  weighing 
3225  lbs.  and  rotating  at  an  angular  velocity  of  300  rev.-per-min., 
the  diameter  of  the  disk  being  4  ft. 

.  3.  Show  that  the  kinetic  energy  of  a  rotating  body  equals 
^Myk"^,  M  denoting  the  mass  of  the  body  and  v^  the  velocity 
of  a  point  of  it  whose  distance  from  the  axis  of  rotation  equals 
the  radius  of  gyration  of  the  body  with  respect  to  that  axis. 

4.  What  is  the  kinetic  energy  of  a  homogeneous  right  cir- 
cular cylinder  which  rolls  so  that  the  speed  of  its  mass-centre 
is  V,  its  mass  being  M?  Ans.  ^Mv"^. 

299.  Potential  Energy  Defined. — A  body  may  possess  energy 
which  is  not  due  to  velocity.  Thus  two  mutually  attracting 
bodies  can  do  work  against  forces  applied  to  either  or  both  if 
allowed  to  move  so  that  they  approach  each  other;  and  as 
mentioned  in  art.  295,  a  compressed  or  stretched  spring  can  do 
work  against  applied  forces  if  permitted  to  resume  its  natural 
length.  The  "change  of  condit^'on  or  state"  in  the  first  case  is 
a  change  in  configuration,  i.e.,  a  change  in  the  positions  of  the 
bodies  relative  to  each  other,  and  in  the  second  case,  if  we  con- 
ceive of  the  spring  as  consisting  of  discrete  particles,  the  change 
is  also  one  in  configuration. 

Energy  of  a  system  of  particles  dependent  on  configuration 
of  the  system  is  called  energy  of  configuration  and,  more  commonly, 
potential  energy. 

300.  The  Amount  of  Potential  Energy  possessed  by  a  system 
in  any  configuration  is  the  work  which  it  can  do  in  passing  from 
that  configuration  to  any  other  taken  as  a  standard,  it  being 
understood  that  no  other  change  of  condition  takes  place.  The 
standard  configuration  maybe  chosen  at  pleasure, but  it  is  con- 
venient to  so  select  it  that  in  all  other  configurations  considered 
the  potential  energy  is  positive. 

Proposition. — The  potential  energy  of  a  system  in  any  con- 
figuration equals  the  amount  of  work  done  by  the  internal  forces 
during  the  change  to  the  standard  configuration. 

Proof :  To  determine  the  potential  energy  we  are  to  compute 
the  work  done  against  the  external  forces  while  the  system 
passes  to  the  standard  configuration,  no  other  change  of  con- 
dition (as  velocity)  of  the  particles  taking  place  so  that  in  the 
passage  to  the  standard  configuration  there  is  no  change  in  the 


3IO 


tVORK  AND  ENERGY. 


[Chap.  XIV. 


kinetic  energy  of  the  system.  It  is  shown  in  art.  306  that  the 
sum  of  the  works  done  bv  the  internal  and  the  external  forces 
during  any  change  of  configuration  equals  the  increment  in  the 
kinetic  energy  of  the  system.  Since  in  the  case  in  hand  there 
is  no  change  in  kinetic  energy,  the  sum  of  the  works  done  by 
the  internal  and  external  forces  equals  zero.  Denoting  the 
internal  and  external  works  by  Wi  and  w^  respectively, 


Wi-\-w^  =  o,     or     Wi  = 


■w^ 


Now  —w^  is  the  work  done  by  the  system  against  the  external 
forces  during  the  passage  to  the  standard  condition,  i.e.,  the 
potential  energy  of  the  system  in  its  initial  configuration ;  hence 
the  last  equation  asserts  the  truth  of  the  proposition. 

301.  Potential  Energy  of  a  System  Not  Always  a  Definite 
Quantity. — The  amount  of  work  done  by  the  internal  forces  dur- 
ing a  change  of  configuration  (and  hence  the  potential  energy 


Fig.  234. 

of  the  system)  may  or  may  not  depend  upon  the  way  in  which 
the  change  of  configuration  takes  place.  This  is  known  to  be 
true  from  direct  experience,  but  it  can  also  be  proved.  Thus, 
let  A,B,  and  C  (fig.  234)  represent  three  bodies  which  may  sUde 
about  on  a  table ;  imagine  them  connected  by  elastic  cords  as 
shown,  and  consider  the  three  bodies,  the  table,  and  the  earth 
as  a  system.  The  cords  are  introduced  merely  as  a  means  of 
applying  certain  forces  to  the  bodies,  but  the  forces  are  to  be 
thought  of  as  exerted  by  one  body  directly  upon  another.  Let 
Aq,  Bq,  and  Cq  represent  a  selected  standard  configuration  of  the 


§11.1  ENERGY.  311 

bodies  and  A,  5,  and  Csome  other  one;  also,  let  F^.Fj,  and  Fgbe 
such  external  forces  acting  upon  them  that  the  passage  to  the 
standard  position  is  without  change  of  velocity. 

Now  the  only  internal  forces  in  the  system  which  do  work 
during  the  motion  are  the  pulls  exerted  by  the  bodies  upon  each 
other  and  the  frictions  (if  any)  between  them  and  the  table. 
As  shown  in  art.  293,  the  work  done  by  the  pulls  exerted  by  any 
two  of  the  bodies  on  each  other  does  not  depend  on  how  the 
final  positions  are  reached.  The  work  done  by  the  frictions, 
however,  does  depend  on  the  manner  of  the  motion;  thus,  if  we 
suppose  the  friction  on  any  body  to  be  constant  in  value,  the 
work  done  by  that  force  equals  the  negative  product  of  the  force 
and  the  length  of  the  path  described  by  that  body.  Hence,  in 
the  system  of  bodies  under  consideration,  the  total  work  done  by 
the  internal  forces  depends  on  the  way  in  which  the  change  of 
configuration  takes  place  and  the  potential  energy  is  not  a 
definite  quantity.  If,  however,  there  is  no  friction,  the  work 
done  by  the  internal  forces  is  independent  of  the  way  in  which 
the  configuration  takes  place,  and  the  potential  energy  is  a 
definite  quantity. 

302.  Conservative  Systems. — If  the  work  done  by  the  internal 
forces  of  a  system  during  any  change  of  configuration  is  inde- 
pendent of  the  way  in  which  the  change  is  made,  the  system  is 
called  conservative,  and  those  internal  force-pairs  (action  and 
reaction)  whose  work  does  not  depend  upon  the  way  in  which 
the  change  takes  place  are  also  called  conservative.  The  poten- 
tial energy  of  such  a  system  in  any  definite  configuration  is  a 
definite  quantity. 

It  is  characteristic  of  conservative  forces  that  they  are  inde- 
pendent of  the  velocity  of  the  particles  on  which  they  act.  We 
consider  only  conservative  forces  which  act  along  the  line  join- 
ing the  particles  between  which  they  act  and  whose  magnitudes 
depend  on  the  distance  between  the  particles.  That  such  forces 
are  conservative  forces  follows  from  art.  293. 

303.  Non-Conservative  Systems. — If  the  total  work  done  by 
the  internal  forces  depends  on  the  manner  of  the  change  of  con- 
figuration, the  system  is  called  non-conservative  and  those  inter- 
nal force-pairs  whose  work  depends  on  the  manner  of  the  change 


312  IVORK  AND  ENERGY.  [Chap.  XIV. 

are  also  called  non-conservative.  The  potential  energy  of  a 
non-conservative  system  in  any  definite  configuration  is  not  a 
definite  quantity. 

It  is  a  characteristic  of  non-conservative  forces  that  their 
magnitudes  or  directions  (or  both)  depend  upon  the  velocity  of 
the  particles  to  which  they  are  applied.  Friction  is  the  only 
one  of  this  class  herein  considered. 

304.  Localization  of  Potential  Energy. — Unlike  the  kinetic 
energy,  the  potential  energy  of  a  system  cannot  always  be 
localized  in  detail,  i.e.,  we  cannot  in  all  cases  assign  to  parts  of 
the  system  certain  definite  parts  of  its  potential  energy.  As  an 
example,  consider  two  bodies  A  and  B  of  the  illustration  in  art. 
301,  and,  for  simplicity,  neglect  friction.  As  previously  ex- 
plained, the  bodies  can  do  a  definite  amount  of  work  against 
external  forces  in  passing  to  their  standard  positions,  A^B^  (fig. 
234),  but  the  amount  of  work  which  each  can  do  depends  upon 
the  way  in  which  the  passage  is  made.  Thus,  suppose  that  they 
move  to  their  standard  positions  successively  and  that  in  one 
passage  A  moves  first  and  in  th6  other  B  moves  first.  Now  the 
work  which  A  can  do  in  each  case  equals  the  work  which  the 
internal  force  acting  on  A  does,  and  this  work  has  different 
values  in  the  two  passages.  For,  let  P  denote  the  internal  force 
acting  on  A  (for  simplicity  assumed  constant) ;  then  (see  ex.  6,  art. 
290)  the  work  done  by  P  in  the  first  case  equals  ±P{Af^B—AB), 
and  in  the  second  case  it  equals  ±P{AqBq  — AB^);  these  values 
are  in  general  unequal. 

If,  however,  one  of  the  bodies  is  always  in  its  standard  posi- 
tion, the  potential  energy  of  the  system  is  rightly  ascribed  to  the 
other.  Thus,  a  magnet  and  a  piece  of  soft  iron  attracting  each 
other  possess  potential  energy  if  separated,  and  if  the  magnet  is 
regarded  as  fixed  the  energy  is  possessed  by  the  iron.  Simi- 
larly, the  earth  and  an  elevated  body  considered  as  a  system 
possess  potential  energy,  but  it  is  practically  necessary  to  regard 
the  earth  as  fixed  and  hence  to  ascribe  the  energy  to  the  elevated 
body. 

305.  Other  Forms  of  Energy. — Kinetic  and  potential  energies 
are  often  called  mechanical  energy.  It  is  the  opinion  of  some 
that   all   energy  is  mechanical,  and  some  think  that  it  is  all 


§11.]  ENERGY,  313 

kinetic.  Whether  either  of  these  views  be  correct,  it  is  practi- 
cally necessary  to  recognize  other  forms.  A  mere  enumeration 
of  these  with  brief  remarks  is  sufficient  for  the  present  purpose, 
since  we  shall  deal  mostly  with  energy  known  to  be  mechanical. 
Thermal  Energy. — A  hot  body  may  do  work  under  favorable 
conditions;  thus,  if  such  a  one  is  placed  in  a  boiler  containing 
water,  the  water  will  be  heated  and  a  part  may  be  converted 
into  steam  which  may  drive  a  steam-engine,  i.e.,  do  work.  By 
giving  up  its  heat  the  hot  body  has  done  work,  and  hence  by 
definition  (art.  295)  it  possessed  energy  in  its  heated  state.  Not 
only  is  this  fact  well  known,  but  also  the  fact  that  a  given  quan- 
tity of  heat  represents  a  definite  amount  of  energy ;  the  relation 
may  be  expressed  thus : 

one  British  unit  of  heat  *  =  778-h  ft. -lbs. 

Based  on  the  molecular  hypothesis,  the  common  theory  is 
that  heat  is  due  to  the  vibratory  motion  of  molecules,  i.e.,  that 
thermal  energy  is  kinetic. 

Chemical  Energy. — Many  substances  combine  chemically  and 
their  combination  gives  evidence  that  they  possessed  energy. 
Thus,  coal  and  oxygen  combine  and  produce  heat  which,  as  "we 
have  seen,  is  a  form  of  energy.  We  rightly  say,  therefore,  the 
coal  and  oxygen  before  combination  possessed  energy. 

Based  on  the  molecular  hypothesis,  the  theory  of  chemical 
energy  in  cases  where  heat  is  generated  in  the  chemical  com- 
bination is  that  internal  (molecular)  forces  of  the  substances 
do  work  during  the  combination,  and  hence  (see  art.  306)  in- 
crease the  kinetic  energy  of  the  molecules.  According  to  this 
explanation  the  energy  before  combination  is  potential  and  after 
kinetic. 

Electrical  Energy. — If  a  storage  battery  charged  with  elec- 
tricity is  connected  with  a  motor,  work  may  be  done  by  the 
latter.  As  the  work  is  done,  the  electrical  condition  of  the  bat- 
tery changes  and  we  therefore  ascribe  the  energy  to  the  battery. 
The  energy  is  called  electrical  because  it  is  due  to  a  change  of 
electrical  condition. 

*  The  amount  of  heat  required  to  raise  the  temperature  of  one  pound 
of  water  one  degree  Fahrenheit. 


314  ,      iVORK  AND  ENERGY.  [Chap  XIV. 

The  nature  of  electrical  energy  is  even  less  understood  than 
that  of  thermal  energy,  and  no  commonly  accepted  explanation 
of  it  has  yet  been  made.  », 

§  III.     Principles  of  Work  and  Energy. 

306.  Principle  of  Work  and  Kinetic  Energy. — I.  For  a  Par- 
ticle.— The  work  done  in  any  displacement  of  a  particle  by  all 
the  forces  applied  to  it  equals  the  increment  in  its  kinetic  energy 
during  that  displacement,  or 

W  =  iE;k (l) 

w  denoting  the  work  done  by  the  forces  and  JE^  the  increment 
in  the  kinetic  energy.  The  increment  is  positive  or  negative 
(and  there  is  a  gain  or  loss  of  kinetic  energy)  according  as  the  final 
velocity  is  greater  or  less  than  the  initial. 

Proof :  Let  R  denote  the  resultant  of  the  forces  acting  on  the 
particle,  w  the  work  done  by  them  in  the  displacement,  v^  and 
V2  the  velocities  of  the  particle  at  the  beginning  and  end  of  the 
displacement  respectively.     Then,  as  shown  in  art.  292, 


IV 


=  r^Rtds, 


where  5  is  the  distance  of  the  moving  particle  from,  some  fixed 
point  in  the  path  and  s^  and  ^2  are  the  values  of  5  at  the  begin- 
ning and  end  of  the  displacement.     Since 

Rt  =  mat  =  m  dv/dt, 
w  =  m  I    ds  dv/dt  =  ml    v  dv  =  imv2^  —  h^Vj^, 

Since  ^mVj^  and  ^mv^^  are  the  values  of  the  kinetic  energy  of 
the  particle  at  the  beginning  and  end  of  the  displacement  re- 
spectively, the  right-hand  member  above  is  the  increment  in 
the  kinetic  energy  of  the  particle. 

II.  For  Any  System  of  Particles. — The  work  done  by  all  the 
external  and  the  internal  fdrces  in  any  displacement  of  a  system 


§111.]  PRINCIPLES  OF  IVORK  AND  ENERGY.  315 

of  particles  equals  the  increment  in  its  kinetic  energy  during  the 
displacement,  or 

w,+Wi  =  JEjfc, (2) 

w^  and  Wi  denoting  the  "external"  and  the  "internal"  works 
respectively. 

Proof:  According  to  the  principle  of  work  and  energy  for  a 
particle  the  total  work  done  on  a  particle  in  any  displacement 
equals  the  increment  in  its  kinetic  energy  during  that  displace- 
ment. Hence  the  work  done  on  all  the  particles  of  a  system 
during  any  displacement  equals  the  sum  of  the  increments  in 
their  kinetic  energies,  i.e.,  the  increment  in  the  kinetic  energy 
of  the  system. 

III.  For  a  Rigid  Body. — The  work  done  upon  a  rigid  body 
by  the  external  forces  in  any  displacement  equals  the  increment 
in  its  kinetic  energy  during  that  displacement,  or 

w,  =  iE, (3) 

Proof:  It  is  shown  in  art.  293  that  the  work  done  by  two 
equal,  opposite,  and  collinear  forces  is  zero  for  any  displacement 
of  their  application  points  if  the  distance  between  those  points 
remains  constant.  As  previously  explained,  the  internal  forces 
of  any  system  of  particles  occur  in  pairs  of  equal,  opposite, 
and  collinear  forces,  and  since  in  a  rigid  body  the  distances 
between  the  application  points  of  the  internal  forces  remain  con- 
stant, the  work  done  by  the  internal  forces  in  any  displacement 
of  the  body  is  zero ;  hence  the  work  done  by  the  external  forces 
equals  the  increment  in  the  kinetic  energy  (see  II). 

307.  Principle  of  Work  and  Energy  for  Conservative  Systems. 
— The  work  done  upon  a  conservative  system  by  external  forces 
during  any  displacement  equals  the  sum  of  the  increments  of 
its  kinetic  and  potential  energies,  or 

w^  =  JEfc  +  iE^, (i) 

AEp  denoting  increment  of  potential  energy. 

Proof:  Let  C^  and  Co  be  the  initial  and  final  and  Cq  the 
standard  configuration  of  the  system.     Also  let  Ep    and  Ep' 


3^^  "  IVORK  AND  ENERGY.  [Chap.  XIV. 

denote  the  potential  energies  in  the  configuration  C^  and  C3 
respectively.  The  work  done  by  the  internal  forces  in  the 
displacement  from  C^  to  Co  would  ha.E/,  and  that  from  C^  to 
Cq  would  be  Ep" ;  hence  in  the  actual  displacement  {C^  to  C^) 
the  work  done  by  the  internal  forces  equals  EJ ~Ep'.  Ac- 
cording to  the  preceding  article, 

w,-^{EJ-E^')=^AEk,     or     w^^AE^-\-{EJ'-Ep'). 

Now  Ep"  —  Ep  is  the  increment  in  the  potential  energy  of  the 
system  during  the  displacement ;   hence  the  principle  is  proved. 

If  the  work  done  by  the  external  forces  is  positive  (that  done 
by  the  system  against  the  .forces  is  negative),  the  system 
gains  energy,  and  if  their  work  is  negative  (that  done  by  the 
system  against  them  is  positive),  the  system  loses  energy. 
The  statement  that  a  certain  amount  of  positive  work  is  done 
upon  (or  by)  a  system  is  equivalent  to  the  statement  that  the 
system  has  gained  (or  lost)  the  same  amount  of  energy. 

308:  Conservation  pf  Energy. — In  any  change  of  condition 
of  a  material  system  which  is  isolated  so  that  it  neither  receives 
nor  gives  out  energy,  its  total  energy  (all  forms  included)  re- 
mains constant  in  amount;  or,  as  is  sometimes  stated,  "energy 
is  indestructible." 

This  principle  is  a  generalization  based  on  physical  experi- 
ence. It  cannot  be  deduced  in  the  general  case  from  the  laws 
of  motion,  at  least  not  in  the  present  state  of  knowledge  regard- 
ing the  constitution  of  matter  and  the  nature  of  non-mechanical 
energy.  For  conservative  systems  the  principle  can  be  proved; 
thus,  the  system  being  isolated,  there  is  no  external  work  and 
equation  (i),  art.  307,  becomes 

JE;fc  +  iE^  =  0, (i) 

i.e.,  the  sum  of  the  increments  of  kinetic  and  potential  energies 
in  any  change  of  condition  equals  zero;  hence  the  sum  of  the 
kinetic  and  potential  energies  is  constant. 

If  a  system  not  isolated  receives  or  loses  energy,  some  other 
system  must  lose  or  receive  an  equal  amount.  For,  let  A  be 
the  first  system  and  B  the  one  from  which  A  receives  or  to  which 


§IIL]  PRINCIPLES  OF  IVORK  AND  ENERGY.  317 

it  gives  energy.  If  necessary,  imagine  B  extended  so  that  A 
and  B  together  are  an  isolated  system;  then  the  total  energy 
of  A  and  B  being  constant,  if  A  receives  or  loses  energy,  B  must 
lose  or  receive  an  equal  amount. 

309.  Principle  of  Energy  for  Machines. — The  function  of  a 
machine  is  to  transfer  energy  from  a  body  or  system  of  bodies 
{A)  to  another  (J5).  The  energy  transferred  may  or  may  not 
have  been  transformed  in  the  process.  Thus,  a  dynamo  re- 
ceives mechanical  and  delivers  electrical  energy,  while  a  water 
motor  receives  and  delivers  mechanical  energy. 

The  energy  received  by  the  machine  from  A  is  called  input 
and  that  delivered  by  it  to  B  is  called  output.  The  amount  of 
energy  possessed  by  the  machine  at  any  instant  is  called  its 
stored  energy  at  that  instant.  It  is  a  fact  of  experience  that 
some  of  the  energy  miscarries,  as  it  were,  between  ^4  and  B,  and 
is  delivered  to  other  bodies  than  B;  that  energy  is  therefore 
called  lost  energy  or  simply  the  loss.  This  energy  is  lost  prin- 
cipally as  heat  which  is  generated  wherever  there  is  a  transfor- 
mation or  transference  of  energy. 

Let  Ei  denote  the  input  for  any  period, 
£^^  the  output. 
El   "   loss, 
AEg    "    increment  in  the  stored  energy; 

then,  since  energy  is  indestructible, 

Ei-{E,  +  Ei)==AE,, 
or  E»  =  E,  +  E,  +  iE« (i) 

If  the  stored  energy  remains  constant,  or  if  at  the  beginning  and 
end  of  the  period  the  stored  energies  are  equal  (as  at  the  begin- 
ning and  end  of  a  cycle  through  which  the  machine  works),  AE, 
equals  zero,  and  the  equation  of  energy  becomes 

E.=  E,+E; (2) 

310.  Efficiency. — The  efficiency  of  a  machine  is  the  ratio  of 
the  output  to  input  for  a  period  at  the  beginning  and  end  of 


31 8  "^        WORK  AND  ENERGY.  [Chap.  XIV. 

which  the  stored  energy  is  the  same.  Thus,  if  e  denotes  effi- 
ciency, 

e  =  E,/E^/ (i) 

Since  the  input  is  always  larger  than  the  output,  the  efficiency 
of  every  machine  is  less  than  one. 

311.  Power  or  Activity. — The  rate  at  which  a  machine  or  any 
"agent"  does  work  is  called  its  power  or  activity. 

If  the  work  is  done  uniformly,  the  power  is  constant;  and  if 
Jw  denotes  the  work  done  in  any  period  Jt  and  P  the  power, 

P  =  Jw/Jt (i) 

If  the  work  is  not  done  uniformly,  the  power  is  variable  and  the 
formula  above  gives  the  average  value  of  the  power ;  the  actual 
value  at  any  instant  is  the  limit  of  Jw/Jt,  or 

P  =  dw/dt •    .     .     .     (2) 

Units  ^pf  Power. — Equations  (i)  and  (2)  imply  as  units  of 
power  a  rate  corresponding  to  unit  work  done  in  a  unit  time. 
Thus  one  foot-pound-per-second,  one  meter-kilogram-per-second, 
one  erg-per-second,  etc.,  are  such  units  of  power.  These  units 
are  small  for  some  purposes;  the  following  are  often  more  con- 
venient :      . 

the  horse-power      =550  foot-pounds-per-second ; 
the  force  de  cheval=    75  meter-kilograms-per-second; 
the  watt  =    10^  ergs-per-second.* 


EXAMPLES. 

1-.  Show  that  the  rate  at  which  steam  does  work  in  any 
engine  is  given  hy  p  I  an,  the  notation  being: 

p,  average  steam-pressure  per  unit  area  during  a  stroke; 
/,  length  of  stroke ; 
a,  area  of  piston ; 
w,  number  of  strokes  per  unit  time. 

*  For  dimensions  of  a  unit  of  power  see  Appendix  C. 


§IV]  APPLICATIONS.  3^9 

2,  Show  that  the  rate  at  which  steam  does  work  in  a  loco- 
motive who^e  velocity  is  v  and  drivers'  diameter  is  d  equals 

4plav/7id.    ■    ,     -^tv-'  '^"//jl/j.- 

§  IV.     Applications. 

312.  Computation  of  Velocity  and  Distance. — The  equa- 
tions expressing  the  principles  of  work  and  kinetic  energy,  (2) 
and  (3),  art.  306, contain  "work  terms"  on  one  side  and  "kinetic 
energy  terms"  on  the  other.  The  factors  in  the  work  terms  are 
force  and  distance,  and  those  in  the  kinetic  energy  terms  are 
mass  and  velocity.  If  all  the  factors  except  one  are  known, 
that  one  may  be  computed  from  the  equation.  In  the  follow- 
ing examples  the  unknown  quantity  is  a  velocity  or  a  distance, 
and  it  can  be  most  readily  determined  by  the  principle  of  work 
and  kinetic  energy. 

EXAMPLES. 

1.  Suppose  that  in  ex.  i,  art.  290,  the  velocity  of  the  body 
5\^hen  at  ^  is  2  ft.-per-sec.  What  is  its  velocity  when  it  reaches 
B  if  the  body  weighs  160  lbs.? 

Solution:  The  total  work  done  by  all  the  external  forces  in 
the  motion  from  A  to  B  was  found  to  be  325  ft.-lbs.  (see  solu- 
tion of  ex.  i).  The  initial  kinetic  energy  of  the  body  (i.e.,  at 
A)  is 

iw2;i^  =  i  4.97X2^  =  9.94  ft.-lbs.; 

hence  eq.  (3),  art.  306,  becomes 

325  =  i4.97i;2^- 9.94,     or     v^  =  \\.(i\  ft./sec. 

2.  Suppose  that  in  ex.  2,  art.  290,  the  velocity  of  the  body 
when  at  A  is  5  ft.-per-sec.  What  is  its  velocity  when  it  reaches 
Bt 

3.  Suppose  that  in  ex.  4,  art.  290,  the  body  weighs  10  lbs. 
and  that  the  horizontal  surface  is  smooth.  What  is  its  velocity 
after  having  moved  10  ft.  ? 

4.  Solve  the  preceding  example  if  the  horizontal  surface  is 
rough,  the  coefficient  of  kinetic  friction  being  one- fourth. 

5.  How  far  will  the  body  of  ex.  4  move  supposing  that  after 


320  IVORK  AND  ENERGY.  [Chap.  XIV. 

the  cord  slackens  the  body  moves  under  the  influence  of  fric- 
tion alone  ?  ♦ 

6.  Suppose  that  in  ex.  4,  art.  266,  the  angular  velocity  of 
the  rotating  body  is  co^  at  a  certain  instant.  Determine  its 
angular  velocity  when  the  suspended  body  has  descended  a  dis- 
tance h  after  that  instant. 

Solution:  The  e"x:ternal  forces  acting  on  the  cord  and  rotat- 
ing and  suspended  bodies  considered  as  a  system  are  the  weights 
of  the  bodies  and  the  "hinge  reaction."  Supposing  the  last  to 
be  frictionless,  it  does  no  work;  neither  does  the  weight  of  the 
rotating  body,  but  the  work  done  by  that  of  the  suspended 
body  is  Wh.  The  work  done  by  forces  internal  to  the  two 
rigid  bodies  and  that  by  the  reactions  between  the  cord  and 
the  two  bodies  is  zero.  The  work  of  the  forces  internal  to  the 
cord  is  not  zero,  but  it  is  small  (see  art.  315)  and  is  negligible  in 
this  instance. 

The  initial  kinetic  energy  of  the  system  (neglecting  the  mass 
of  the  cord)  is 

I  W 
^  s 
and  if  co^  denotes  the  final  angular  velocity,  eq.  (2),  art.  3061 
becomes,  since  a>,  =  o, 

from  which  (O2  can  be  computed. 

•  7.  Suppose  that  the  disk  of  ex.  2,  art.  298,  is  rotating  on  a 
shaft  4  in.  in  diameter  and  that  the  axle  friction  is  30  lbs.  In 
how  many  turns  would  the  frictional  resistance  stop  it  ? 

8.  Let  h  denote  the  height  of  the  centre  of  gravity  of  a  pen- 
dulum when  the  velocity  is  zero  above  its  lowest  position.  Show 
that  the  angular  velocity  of  the  pendulum  when  it  reaches  its 
lowest  position  is  {i/k)\^2gh,  k  denoting  the  radius  of  gyration 
of  the  pendulum  with  respect  to  the  axis  of  suspension. 

9.  A  body  is  suspended  by  two  parallel  cords  of  equal  length 
and  is  allowed  to  swing  in  the  plane  of  the  cords  under  the  in- 
fluence of  gravity.  Show  that  the  height  (h)  to  which  the  cen- 
tre of  gravity  rises  above  its  lowest  position  and  the  velocity  (v) 
in  the  lowest  position  are  related  as  follows :  v"^  =  2(i}i. 


§IV.]  APPLICATIONS,  321 

313.  Train  Resistance. — The  resistance  to  motion  experi- 
enced by  a  train  moving  on  a  track  consists  of  rolling  resistance, 
journal  friction  (at  the  car  and  locomotive  axles),  air  resistance, 
and  the  frictional  resistance  at  the  working  parts  of  the  loco- 
motive. For  simplicity  we  may  imagine  these  replaced  by  a 
single  horizontal  resistance  equivalent  to  them  all  so  far  as  the 
motion  of  the  train  is  concerned.  This  single  equivalent  resist- 
ance is  often  called  train  resistance^  but  sometimes  the  term  is 
intended  not  to  include  the  locomotive  and  tender  resistance. 

It  is  customary  to  express  train  resistance  as  so  many  pounds 
for  each  ton  of  weight  of  train.  Many  experiments  have  been 
made  to  determine  train  resistance,  and  a  number  of  formu- 
las have  been  deduced  to  express  its  value.  We  give  but  three 
as  illustrations  and  to  furnish  data  for  a  few  examples;  the 
notation  is  as  follows: 

R,  train  resistance  in  Ibs.-per-ton; 

F,  velocity  in  mi.-per-hour; 

r,  weight  of  train  in  tons  (2000  lbs.). 

Engineering  News :  R  =  V/^-\-2 (i) 

This  includes  all  resistances  except  the  internal  friction 
of  the  locomotive.  For  the  higher  velocities  it  was  deduced 
from  experiments  on  a  fast  passenger-train. 

Crawford:  R  =  Vy26%  +  2.s (2) 

This  does  not  include  resistance  of  locomotive  and  tender. 
It  was  deduced  from  experiments  on  passenger-trains. 

Lundie:  i^  =  4 +  VT0.2  + 14/(35 +  7^)] (3) 

This  was  deduced  from  experiments  on  electric  elevated 
street-railway  trains  and  includes  all  resistances. 
It  will  be  noticed  that  only  (3)  contains  T.     The  first  two 

apply  only  to  trains   of  approximately   the   same  kind   and 

weight  as  those  experimented  on. 


$22  "  1VORK  AND  ENERGY.  [Chap.  XIV. 

EXAMPLES. 

V   I.  Plot  the  curves  represented  by  the  preceding  equations, 
assuming  T  in  the  third  to  be  one  ton. 

%^  2.  What  is  the  relation  between  the  "draw-bar  pull"  be- 
tween tender  and  first  car  when  the  velocity  of  the  train  is  con- 
stant ? 

Solution:  The  only  forces  doing  work  on  the  train  are  the 
draw-bar  pull  and  the  train  resistance.  Since  the  velocity  of 
the  train  is  constant,  its  kinetic  energy  remains  constant  and 
the  total  work  done  by  the  two  forces  must  be  equal  to  zero 
(see  eq.  (2),  art.  306),  i.e.,  the  works  done  by  them  must  be 
equal  but  of  opposite  sign.  Since  the  forces  act  through  equal 
distances  in  any  motion  of  the  train,  the  forces  must  be  equal 
in  amount. 

Train  resistance  (for  the  cars)  is  determined  by  measuring 
the  draw-bar  pull  in  front  of  the  first  car  when  the  velocity  is 
constant. 

3.  Assume  the  train  resistance  for  cars  to  be  constant  (as  it 
is  roughly  below  10  mi.-per-hr.,  according  to  Crawford's  for- 
mula) and  that  it  equals  2.6  Ibs.-per-ton.  If  the  cars  with  load 
weigh  1000  tons,  how  much  work  does  the  engine  do  upon  the 
cars  to  bring  up  the  velocity  of  the  train  from  o  to  10  mi.-per-hr., 
if  it  is  done  in  a  distance  of  one  mile  on  a  level  track  ? 
{-  4.  Solve  the  preceding  example  supposing  that  the  train 
runs  up  a  "  one-per-cent  grade."  * 

5.  Supposing  that  the  speed  in  ex.  3  is  increased  uni- 
formly, i.e.,  the  acceleration  is  constant,  what  is  the  value  of 
the  draw-bar  pull? 

6.  Express  the  rate  at  which  the  engine  does  work  on  the 
train  in  the  preceding  example  in  horse-powers  when  the  train 
is  starting  and  when  its  velocity  is  10  mi.-per-hr. 

^  7.  Show  that  if  P  denotes  the  rate  at  which  work  is  done  by 
the  steam  in  a  locomotive,  i.e.,  the  rate  at  which  energy  is  sup- 
plied to  the  engine,  R  the  total  train  resistance,  and  v  the  speed, 
when  V  is  constant,  P  =  Rv. 

*  One  ft.  rise  for  every  100  ft.  along  the  track. 


§  IV.]  APPLICA  TIONS.  3  2  3 

Solution:  The  speed  being  constant,  there  is  no  change  in 
the  kinetic  energy  of  the  train ;  therefore  the  work  done  on  the 
engine  by  the  steam  equals  the  work  done  by  the  train  against 
the  resistance  during  any  period.  If  w  denotes  the  first  work 
during  which  the  train  moves  any  distance  s,  in  a  time  t, 

w  =  Rs,     or     w/t^Rs/t. 

Now  w/t  is  the  rate  at  which  work  is  done  on  the  engine,  and 
s/t  is  the  velocity  of  the  train;  hence  P  =  Rv. 

8.  If  the  Engineering  News  formula  is  correct  at  all  speeds 
and  gives  practically  the  whole  train  resistance,  show  that  the 
cylinder  steam-pressure  per  unit  area  (average  for  one  stroke) 
required  to  maintain  any  constant  speed  in  a  train  on  a  level 
track  is  a  linear  function  of  the  speed.     (See  ex.  2,  art.  311.) 

9.  Show  that  when  the  speed  of  a  train  is  changing,  P  = 
Rv-\-mva,  m  and  a  denoting  the  mass  and  acceleration  of  the 
train  respectively  and  P,  R,  and  v  having  meanings  as  in  ex.  7. 
(The  equation  neglects  the  "rotational"  component  of  the 
kinetic  energy  of  the  wheels,  which  is  small  compared  to  the 
kinetic  energy  of  the  train.) 

314.  Friction  Brakes. — Fig.  235  represents  a  form  of  friction 

brake    often   used    to    measure   the  power  of         ^ ^ 

small  motors.     It  consists  essentially  of  a  fiat-      f  \ 

faced  pulley  rigidly  fastened  to  the  shaft  of      [         c<-— -r- 

the  motor,  a  strap  or  rope  partly  encircling  the      K  J 

pulley,   one   end    of   it    being    fastened  to    a     fli^^^__^^ 
spring-balance  and  the  other  sustaining  a  freely     m  LJ 

hanging  body.  z£7  ^ 

When  the  pulley  rotates  it  drags  the  strap  Fig.  235. 
around  with  it  a  small  distance  and  the  spring  tension  (T)  is 
greater  or  less  than  W  according  as  the  rotation  is  clockwise  or 
counter-clockwise.  Assuming  it  to  be  clockwise,  the  frictional 
resistance  at  the  rim  of  the  wheel  equals  T  —  W  and  the  work 
done  by  the  motor  against  friction  in  one  revolution  of  the 
wheel  is  {T  —  W)27tr.  Hence,  if  the  wheel  makes  n  revolutions 
per  unit  time  and  all  of  the  work   done  by  the  motor  is  thus 


324 


^f^ORK  AND  ENERGY. 


[Chap.  XIV. 


expended    against    friction,    the   power   of   the   motor   equals 

{T  —  W)27:rn. 

Fig.  236  represents  a  form  of  brake  sometimes  used  on  hoist- 
ing-drums for  stopping  the  same. 
It  consists  essentially  of  a  strap 
partially  encircling  the  drum  or  a 
wheel  fastened  to  the  drum,  one 
end  being  fastened  to  a  fixed  sup- 
port, as  A,  and  the  other  end  to  a 
lever,  as  CD.  A  force  (P)  applied 
as  shown  brings  a  frictional  resist- 
ance to  bear  upon  the  drum  at  the 
strap  and  so  controls  the  speed.  We 
wish  to  find  the  relation  between 

the  pressure  P  and  the  weight  of  the  descending  load  when  its 

velocity  is  constant. 

Let  Ti  and  Tj  denote  the  tensions  in  the  strap  at  A  and  B 

respectively,  F  the  frictional  resistance  at  the  brake-strap,  /  the 

coefficient  of  friction,  F'  the  axle  friction,  and  /'  the  coefficient 

of  axle  friction.     Since  the  kinetic  energy  of  the  moving  system 

is  constant, 

W2nr  =  F2na-^F'27tc (i) 


Fig.  236. 


A-S  shown  in  art.  r6o, 


T^ef^ 


(2) 


Considering  the  forces  on  the  lever,  it  is  seen  that 


Also, 
and 


T^d  =  Pb.  .  .  . 
F'  =  r(W  +  T,  +  T,) 
F=f^-T^.    .     .     . 


(3) 

(4) 
(5) 


These  five  equations  determine  all  the  unknown  quairtities  P, 
Ti,  r„  F,  and  F'. 

315.  Efficiency  of  Tackle. — Fig.  237  represents  a  fixed  pul- 
ley. Let  a  denote  its  radius,  r  that  of  the  axle,  and  d  the  diam- 
eter of  the  rope ;  also  let  P  and  Q  denote  the  tensions  on  the  two 
sides  as  shown. 


§IV.] 


APPLICATIONS. 


325 


Consider  the  pulley  and  as  much  of  the  rope  as  is  shown 
together  as  a  machine.  In  any  motion,  the  external  works  are 
those  done  by  P,  Q,  and  the  axle  friction;  the  internal  work  is 
done  by  friction  between  the  rope  fibres  where  the  rope  winds 
on  and  off.  In  one  revolution  of  the  wheel  the  works  done  by 
P  and  Q  respectively  are  Pina  and  —  Q27ra, 
and  that  done  by  the  axle  friction  is—j'Rinr, 
/'  denoting  the  coefficient  of  axle  friction  and 
R  the  resultant  axle  pressure.  Considering 
that  numerical  values  of  /'  are  uncertain,  we 
may  write  R  =  2Q;  then  the  work  of  axle 
friction  is  approximately  —j'Q^nr.  It  has 
been  found  experimentally  that  the  work 
"due  to  rigidity"  of  the  rope  is  proportional 
to  Q,  i.e.,  to  the  tension  on  the  following 
side;  hence  we  may  write  for  this  work  in 
one  revolution  —cQ27cr,  c  being  an  experimental  coefficient. 

The  equation  of  work  and  energy  is  (for  one  revolution) 


Fig.  237. 


or 


P27:a  —  Q2na—fQ^nr—Q27tcr  =  o, 


(I) 


k  being  an  abbreviation  for  (i+cr/a+2/V/a),  sometimes  called 
**  coefficient  of  resistance  "  for  the  pulley. 

With  equation  (i)  we  can  compute  the  relation  between  the 
load  and  the  necessarj^  pull  to  raise  or  lower  it,  and  the  efficiency 
of  the  tackle  (see  ex.  1).  The  value  of  the  coefficient  c  has  been 
found  to  vary  directly  as  the  square  of  the  diameter  of  the  rope 
{d)  and  inversely  as  the  radius  of  the  pulley.  One  formula 
(Eytelwein's)  is 


d  and   a  being   expressed   in   inches. 
f  =  0.1,  the  values  of  k  are  as  follows: 


If  a  =  4C?,  r—d/2,  and 


d= 

h 

f 

I 

li 

k  = 

1.08 

I .  II 

1. 14 

1 .20 

326  "       H^ORK  AND  ENERGY.  [Chap.  XIV. 

EXAMPLES. 

1.  Determine  the  relation  between  F  and  W  (fig.  83)  and 
the  efficiency  when  the  load  is  being  raised. 

Solution :  Call  the  tensions  in  the  sections  made  by  the  hori- 
zontal line  beginning  at  the  left  5i,  Sg,  Sg,  etc.  Then  S2  =  kSir 
S^  =  k^Si,  S^^k^Si,  etc.,  and  since 

l^  =  5i  +  52+53+ etc., 

Also  F  =  kSQ  =  k^Si', 

hence  F^^Wk'^/ii+k  +  k^+k^+k^+k^). 

During  an  ascent  of  the  load  equal  to  s,  the  point  of  appli- 
cation of  F  descends  a  distance  65.  Hence  for  that  motion  the 
output  and  input  are  respectively 

Ws    and    6Fs, 
anu  the  efficiency  equals 

W/6F=(i+k+k^-\-  .  .  .  k^)/6k\ 

2.  Solve  the  preceding  example  supposing  that  the  load  is 
being  lowered. 

316.  Efficiency  of  a  Mine-hoist. — Fig.  238(a)  represents  a 
balanced  vertical  mine-hoist,  consisting  of  an  endless  cable, 
two  cages ,  a  hoisting-drum  above,  and  a  pulley  below.  Ordinates 
from  Ot  to  the  various  lines-  in  fig.  238(6)  represent  various 
quantities  involved  in  a  power  computation  which  is  now  to  be 
made.  Thus,  ordinates  to  aa  represent  a  frictional  resistance 
(2  tons)  applied  to  the  surface  of  the  drum  equivalent  to  all  the 
actual  frictional  resistances  in  the  hoist;  ordinates  to  bb  repre- 
sent the  weight  of  the  load  (6  tons) ;  and  ordinates  to  cc  repre- 
sent the  velocities  of  the  cages.  The  assumed  law  of  motion  is 
(i)  that  the  acceleration  is  greatest  at  the  beginning  of  the 


§IV.] 


APPUCATIOm. 


327 


motion  and  decreases  to  zero  uniformly  and  so  that  the  velocity 
acquired  in  30  sees,  is  60  ft.-per-sec,  (2)  then  the  velocity  re- 
mains constant  for  60  sees.,  (3)  then  a  retardation  follows  which 
increases  just  as  the  acceleration  decreased,  i.e.,  uniformly  and 
so  that  the  cages  are  brought  to  rest  in  30  sees. 

When  the  speed  of  the  cages  is  constant  the  pull  (F)  of  the 
hoisting-engine  (assumed  for  simplicity  as  applied  to  the  sur- 
face of  the  drum)  just   equals  the  sum  of  the  load  and  the 


100  ft.  per  sec. 


(6) 


Fig.  238. 


frict^'on,  but  in  the  first  30  sees,  the  pull  is  increased  by  the 
"inertia  force,'*  and  in  the  last  30  sees,  it  is  decreased  by 
the  "  inertia  force."  This  inertia  force  depends  upon  the 
mass  of  the  cable,  load,  and  cages,  the  moment  of  inertia 
of  the  drum,  and  the  acceleration  or  retardation.  We 
suppose  that  these  are  such  that  the  ordinates  to  dd'  and 
to  d''d  represent  the  values  of  the  force  as  applied  to  the 
surface  of  the  drum  and  that  its  maximum  value  is  12  tons. 
The  total  force  then  to  be  exerted  by  the  engine  on  the  drum  at 
any  instant  is  represented  by  the  sum  of  the  ordinates  to  the 
three  lines  aa,  bb,  and  dd  for  that  instant.  Ordinates  to  the 
line  ee  represent  all  such  sums. 

The  power  of  the  engine  P  at  the  instant  when  the  velocity 
is  V  and  the  work  done  by  it  if  in  a  distance  5  or  a  time  t  from 
starting  are  given  by 

P  =  Fv,     w=  r  Fds,     w=  f^  Pdt. 

Jo  Jo 


328  ,      H^ORK  AND  ENERGY,  [Chap.  XIV. 


EXAMPLES. 


1.  Draw  a  curve  showing  how  the  •power  changes  with  the 
time. 

2.  Determine  the  entire  work  done  by  the  engine  in  a  single 
hoist,  and  also  the  efficiency  of  the  hoist. 


CHAPTER  XV. 

IMPULSE  AND  MOMENTUM. 

§  I.     Impulse. 

317.  Impulse  of  a  Force  whose  Direction  is  Constant. — If 

the  magnitude  of  the  force  is  constant,  the  impulse  of  the  force 
for  any  interval  is  the  product  of  the  force  and  the  length  of  the 
interval,  i.e.,  if  F  and  {t"  —  t')  denote  the  force  and  the  interval 
respectively,  the  impulse  equals  F(t"  — t'). 

If  the  magnitude  of  the  force  varies,  the  impulse  for  any  ele- 
ment of  time  equals  the  product  of  the  value  of  the  force  at  any 
instant  of  the  element  and  the  length  of  the  element,  i.e.,  Fdt\ 

and  the  impulse  for  any  finite  interval  t"  —  t'  equals  /  ^  Fdt. 

The  unit  of  impulse  depends  on  the  units  used  to  express  force 
and  time.  If  C.G.S.  units  be  used,  the  unit  of  impulse  is  called 
a  dyne-second;  if  the  pound  and  second  are  used  as  units  of  force 
and  time  respectively,  the  unit  of  impulse  is  called  a  pound- 
second.* 

318.  Impulse  of  a  Force  whose  Direction  Varies. — An  impulse 
should  be  regarded  as  a  vector  quantity,  its  direction  being  the 
same  as  that  of  the  force  if  that  is  constant. 

The  meaning  of  impulse  of  a  force  whose  direction  varies  may 
be  explained  as  follows :  Imagine  a  force  to  change  its  direction 
and  magnitude  five  times  in  a  certain  interval,  and  that  the 
values  of  the  force  and  the  corresponding  portions  of  the  inter- 
val are  F',  F" ,  etc.,  and  {M)' ,  {At)",  etc.,  respectively.  Then 
the  impulses  of  the  force  for  these  portions  are  F'{At)\  F"{Jty\ 
etc.  Now  if  Aa,  ah,  etc.  (fig.  239),  represent  these  impulses,  the 
impulse  of  the  force  for  the  entire  interval  is  the  vector  sum  of 
Aa,  ah,  etc.,  or  AB. 

*  For  dimensions  of  a  unit  impulse  see  Appendix  C. 

329 


330  fMPULSE  AND  MOMENTUM.  [Chap.  XV, 

If  the  force  changes  by  small  amounts  and  many  times  in 

the  interval,  its  variation  may  resemble  that  of  a  continuously 

varying  force.     By  impulse  of  a  conti!iuously  varying  force  is 

meant  the  limit  toward  which  the  impulse  of  a  suddenly  varying 

c  force  tends  as  the  manner  of  varia- 

tion  of  the  latter  approaches  that  of 

X  B       the  former. 

j  Employing  the  language  of  the 

I         infinitesimal  calculus  again,  we  state 
j         that  the  impulse  of  a  force  for  an  ele- 
j^       ment   of   time  equals  F-dtii  F  de- 
notes the  value  of  the   force  at  any 
^  ^    instant  of  the  element,  and  its  direc- 

tion  is  that  of  the  force.     The  im- 
pulse of  the  force  for  a  finite  interval  is   /  Fdt,  the  integration 

being  not  ordinary  but  vectorial.* 

319.  Component  of  an  Impulse. — Since  an  impulse  is  a  vector 
quantity  it  can  be  resolved;  thus  let  AB  (fig.  239)  represent  an 
impulse  and  OX  an  x  axis ;  then  the  x  component  of  the  im- 
pulse is  represented  by  AC. 

Proposition. — The  component  of  the  impulse  of  a  force  along 
any  line  equals  the  impulse  of  the  component  of  the  force  along" 
that  line. 

Proof:  Let  F  denote  the  value  of  the  force  at  any  instant  and 
a  its  angle  with  the  line  (the  x  axis,  say).  Then  for  any  element 
of  time  {dt)  including  the  instant,  the  x  component  of  the  im- 
pulse of  the  force  is  {Fdt)  cos  a.  Evidently  the  x  component  of 
the  impulse  for  a  finite  interval  t"  —  f  equals  the  sum  of  the  x 

Gompotients  of  the  elementary  impulses,  i.e.,  /    Fdt-co^  a.     But 
f  Fdt- cos  a=  f  Fa-dt, 

Fx  denoting  the  x  component  of  F.     Since  the  second  integral 

*  While  the  student  may  not  be  able  to  compute  the  impulse  of  a 
force  in  this  way,  it  is  desirable  that  he  should  understand  the  principle 
of  the  method  as  above  given. 


il.] 


IMPULSE. 


331 


is  the  impulse  of  the  x  component  of  the  force  for  the  interval, 
the  proposition  is  proved. 

320.  Moment  of  the  Impulse  of  a  Force. — We  regard  the  im- 
pulse of  a  force  as  having  not  only  magnitude  and  direction,  but 
also  "position."  If  the  action  line  of  the  force  is  fixed,  then 
that  line  is  also  the  position  line  of  the  impulse.  If  the  action 
line  changes,  then  the  position  line  of  the  impulse  for  an  element 
of  time  coincides  with  the  action  line  of  the  force  at  any  instant 
of  the  interval. 

I.  The  Force  is  Constant  and  its  Action  Line  is  Fixed. — Let 
ah  (fig.  240)  represent  such  a  force  (F)  and  OX  an  axis  of  mo- 
ments. The  impulse  of  the  force  for  a 
period  t"  —  t'  is  F{t"  —  t'),  its  position  line 
being  ab. 

The  moment  of  the  force  about  OX  is 
defined  in  art.  28  as  the  product  of  the 
component  of  the  force  which  is  perpen- 
dicular to  OX  (the  other  being  parallel  to 
OX)  and  the  distance  between  the  perpen- 
dicular component  and  the  axis.  That  is,  if  M  denotes  the 
moment  and  p  the  distance, 

M  =  {F  sin  a)p ^    .     (i) 

In  an  analogous  way  we  define  the  moment  of  the  impulse  of 
the  force  (or  the  angular  impulse  of  the  force,  as  it  is  also  called) 
to  be  the  product  of  that  component  of  the  impulse  which  is 
perpendicular  to  the  axis  and  the  distance  between  that  com- 
ponent and  the  axis.  If  ab  is  taken  to  represent  the  impulse, 
ac  represents  the  perpendicular  component,  and  the  angular  im- 
pulse equals 

F(t"-tOsina-p  =  M(t"-t'),     ....     (2) 


i.e.,  for  any  period  the  moment  of  the  impulse  about  any  axis 
of  a  constant  force  whose  action  line  is  fixed  equals  the  product 
of  the  moment  of  the  force  about  that  line  and  the  length  of 
the  period. 

II.  The  Force  Varies  in  Any  Way. — The  moment  of  the  im- 
pulse about  any  line  for  an  element  of  time  is  the  product  of  the 


332  IMPULSE  AND  MOMENTUM,  [Chap.  XV. 

value  of  the  moment  of  the  force  about  that  line  at  any  instant 
of  the  element  and  the  element,  i.e.,  Mdt.     The  moment  of  the 

impulse  for  a  finite  interval  t"  —  t'  is  the  sum  of  the  angular  im- 

Xt" 
,  Mdt. 

The  Unit  of  an  Angular  Impulse  is  the  angular  impulse 
of  a  force  whose  impulse  is  such  that  its  component  perpendic- 
ular to  a  moment  axis  equals  a  unit  impulse  and  has  an  arm  of 
unit  length.  There  are  no  names  in  use  for  these  units.  To 
describe  the  unit  of  any  numerical  value  of  an  angular  impulse, 
or  moment  of  an  impulse,  we  name  the  units  of  impulse  and 
length  used  in  the  computation. 

The  rule  of  signs  for  moments  of  impulses  is  like  that  for 
moments  of  forces  (see  art.  28),  i.e.,  we  give  the  same  sign  to  the 
moment  of  an  impulse  of  a  force  with  respect  to  an  axis  as  we 
give  to  the  moment  of  the  force  with  respect  to  that  axis. 

§  II.     Momentum. 

321.  Momentum  of  a  Particle. — The  momentum  of  a  particle 
is  the  product  of  the  mass  (m)  and  the  velocity  {v)  of  a  particle. 

Unit  of  Momentum. — The  definition  implies  as  unit  the  mo- 
mentum of  a  particle  of  unit  mass  moving  with  unit  velocity. 
The  magnitude  of  the  unit  hence  depends  upon  the  units  of  mass 
and  velocity  employed.  No  single  words  have  been  generally 
accepted  as  names  for  any  units  of  momentum.  It  is  shown  in 
Appendix  C  that  the  dimensions  of  a  unit  momentum  are  the 
same  as  those  of  a  unit  impulse;  hence  it  is  not  inappropriate 
to  call  these  units  by  the  same  name.  Thus  in  the  C.G.S.  sys- 
tem the  unit  of  momentum  is  called  a  dyne-second,  in  the 
English  engineers'  system  it  is  called  a  pound  (force)-second,  etc. 

322.  Components  of  a  Momentum. — Momentum  should  be 
regarded  as  a  vector  quantity,  its  direction  being  the  same  as 
that  of  the  velocity  of  the  particle.  Like  any  other  vector  quan- 
tity, a  momentum  may  be  r-esolved;  thus  if  od  (fig.  241a)  rep- 
resents the  velocity  of  a  particle,  to  some  scale  it  also  represents 
the  momentum  {mv),  and  oa  and  oa'  represent  two  components 
of  the  momentum,  their  values  being  mv  cos  a  and  rnvsina 


§11.] 


MOMENTUM. 


respectively.     Also  oa,  ob,  and  oc  sue  the  x,  y,  and  z  compo- 
nents of  the  momentum,  and  their  values  are 

mv  cos  a  =  mv^,     mv  cos  ^  =  mv^,     mv  cos  ;-  =  mv^. 

323.  Moment  of  Momentum. — Momentum  should  be  regarded 
as  having  not  only  magnitude  and  direction,  but  also  "position.* 


a 

'^^ 

c 
0 

IV 

X. 

(a) 


a>) 


Fig.  241. 


The  tangent  line  along  which  the  velocity  at  any  instant  is 
directed  is  the  position  line  of  the  momentum. 

Analogous  to  the  moment  of  a  force  with  respect  to  an  axis 
(art.  28)  we  define  the  moment  of  momentum  of  a  particle  (or 
angular  momentum,  as  it  is  also  called)  to  be  the  product  of  that 
component  of  the  momentum  which  is  perpendicular  to  the  axis 
(the  other  being  parallel  to  it)  and  the  distance  between  the  per- 
pendicular component  and  the  axis.  Thus,  the  momentum  rep- 
resented by  od  (fig.  241a)  has  a  moment  about  the  line  OX  equal 
to  oa'  Xp,  p  denoting  the  distance  from  the  axis  to  oa\ 

The  unit  of  angular  momentum  is  the  angular  momentum 
of  a  particle  whose  momentum  is  such  that  its  perpendicu- 
lar component  equals  a  unit  momentum  and  has  a  unit  "  arm.'* 
There  are  no  names  in  use  for  these  units.  To  describe  the  unit 
of  any  numerical  value  of  a  moment  of  momentum  we  name  the 
units  of  momentum  and  length  used  in  the  computation. 

The  rule  of  signs  for  moments  of  momentum  is  similar  to 
that  for  moments  of  forces  (art.  28).  We  imagine  the  "perpen- 
dicular component"  to  be  a  force  and  then  the  sign  of  the 


334  "- IMPULSE  AND  MOMENTUM.  [Chap.  XV. 

moment  of  momentum  is  like  that  of  the  moment  of  that  force. 
Thus,  in  the  preceding  illustration,  the  angular  momentum 
about  the  x  axis  is  positive.  * 

Proposition. — If  the  momentum  of  a  particle  be  resolved  into 
three  rectang-ular  components,  the  moment  of  momentum  with 
respect  to  a  line  parallel  to  one  of  the  components  equals  the 
sum  of  the  moments  of  the  other  two  components  with  respect 
to  that  line.     (Compare  Prop.  I,  art.  28.) 

Proof:  Let  od  (fig.  241)  represent  the  momentum,  OX,  OK, 
and  OZ  directions  of  resolution,  and  OX  the  moment  axis.  As 
explained  in  the  foregoing,  the  moment  of  momentum  with 
respect  to  that  axis  is 

oa^Xp  =  (mv  sin  a)p. 

From  fig.  241(6),  which  represents  the  lines  in  the  plane  boc  in 
their  true  relations,  it  is  plain  that 

p  =  y  cos  f'  —  z  sin  ;-' ; 
hence  the  moment  of  momentum  equals 

mv  sin  a  {y  cos  f  —z  sin  ;-'), 
or  {mvz)y  —  {'mvy)z. 

324.  Momentum  of  a  System  of  Particles. — By  momentum  of 
a  system  of  particles  is  meant  the  resultant  *  of  the  momenta 
of  the  particles.  We  shall  not  need  a  general  expression  for  this 
resultant,  but  will  deduce  the  value  of  its  component  along  any 
line  and  the  value  of  the  resultant  in  special  cases  (art.  326), 

Just  as  in  a  system  of  forces,  the  component  of  the  resultant 
of  any  number  of  momenta,  along  a  line  equals  the  algebraic 
sum  of  the  components  of  the  momenta  along  that  line.  Thus 
the  X  component  of  the  resultant  equals,  if  m',  w",  etc.,  denote 
the  masses  of  the  particles  and  ^'^  v",  etc.,  their  velocities, 

♦wW  +  w'W'+  •  .  .  =ImVx=Mvg  (see  eq.  (2),  art.  239), 

♦  Computed  according  to  the  methods  for  compounding  forces  as 
given  in  Chap.  II. 


§  II.]  MOMENTUM.  335 

M  being  the  mass  of  the  system  and  v  the  velocity  of  its  mass- 
centre.  That  is,  the  component  of  the  momentum  of  a  system 
of  particles  along  any  line  is  the  same  as  if  the  entire  mass  were 
concentrated  at  the  mass-centre. 

325.  Moment  of  the  Momentum  of  a  System  of  Particles. — 
By  moment  of  momentum  of  any  system  of  particles  about  any 
line  is  meant  the  algebraic  sum  of  the  moments  of  the  momenta 
of  the  particles  about  that  line. 

As  explained  in  art.  323,  the  moment  of  the  momentum  of  a 
particle  whose  mass,  velocity,  and  coordinates  are  m',  v' , 
{x\  y,  z')  about  the  x  axis  is 

{m'v/y'  —fn'VyZ)\ 

hence  the  moment  of  momentum  of  the  system  about  the  x  axis 
is 

^(mv^y-mVj/Z). 

326.  Momentum  of  a  Rigid  Body  in  Special  Cases. — I.  A 

Translating  Body. — The  velocities  of  all  particles  being  the  same 
in  magnitude  and  direction,  the  momentum  equals,  if  v  denotes 
the  common  velocity,  (dm)^,  {dni)^,  etc.,  the  masses  of  the  par- 
ticles, and  m  the  mass  of  the  body, 

(dm)iV  +  (dm)2V  +  .  .  •  =vi'dm  =  mv, 

and  the  direction  of  the  momentum  is  the  same  as  that  of  the 
velocity. 

The  position  line  of  the  momentum  contains  the  mass-centre, 
as  can  be  shown  by  a  method  similar  to  that  employed  in  art. 
244.      Hence    the   moment    of   the 
momentum    about   any  line  is   the 
same  as  if  the  entire  mass  were  con- 
centrated at  the  mass-centre. 

11.  A  Rotating  Body. — We  assume 

as  in  art.  262  that  the  rotating  body 

is  homogeneous  and  has  a  plane  of 

symmetry  perpendicular  to  the  axis 

c       .   ^'  J    .    n  .  Fig.  242. 

of  rotation.     Let  fig.  242  represent 

that  section  of  symmetry,  C  the  mass-centre,  and  0  the  centre  of 

rotation.     As  in  art.  242,  imagine   the  body  divided  into  ele- 


I 


33^  *"  IMPULSE  AND  MOMENTUM.  [Chap.  XV. 

mentary  rods  parallel  to  the  axis  of  rotation.  Evidently  the 
position  line  of  the  momentum  of  each  rod  is  in  the  plane  of 
symmetry ;  hence  that  of  the  momeiltum  of  all  the  rods  (or  the 
body)  must  be  in  that  plane.  Since  the  components  of  the 
momentum  of  the  body  parallel  to  the  x  and  y  axis  equal  mvx. 
and  niuy  respectively  (see  art.  324),  the  resultant  momentum 
equals 

{{mVx)  ^  +  {'yii^y)  ^]  =  ^^  —  mroj 

{oj  denoting  the  angular  velocity  of  the  body),  and  its  direction 
is  the  same  as  that  of  the  velocity  of  the  mass-centre.  The  posi- 
tion line  of  the  momentum  passes  through  a  point  Q  in  the  line 
OC,  whose  distance  {q)  from  the  axis  of  rotation  is  given  by 

k  being  the  radius  of  gyration  of  the  body  with  respect  to  the 
axis  of  rotation.  This  fact  can  readily  be  proved  from  the 
result  of  the  remainder  of  this  article. 

The  moment  of  the  momentum  about  the  axis  of  rotation 
might  be  computed  from  art.  325,  but  the  following  method  is 
as  simple  and  preferable:  Let  P(fig.  242)  represent  any  particle, 
dm  its  mass,  and  r  its  distance  from  the  axis  of  rotation;  then 
the  momentum  of  that  particle  is  dmrco,  its  position  line  as 
shown,  and  the  moment  of  its  momentum  equals  {dmroj)r. 
Hence  the  moment  of  the  momentum  of  the  body  equals 


/  dmr'^oj  =  ajj  dm  •  r^  =  Icu, 


I  denoting  the  moment  of  inertia  of  the  body  with  respect  to 
the  axis  of  rotation.  This  resultant  is  independent  of  the  as- 
sumption made  in  the  first  paragraph. 

§  III.     Principles  of  Impulse  and  Momentum. 

327.  Principles  for  a  Particle. — Let  a,  v,  and  {x,  y,  z)  denote 
the  acceleration,  velocity,  and  coordinates  at  any  instant  of  a 
particle  whose  mass  is  m,  and  let  R  denote  the  resultant  of  all 
the  forces  applied  to  it.     Then,  according  to  art.  236, 

Rx  =  ma^  =  mdvg/dt, 
or  Rfdi  =  mdvx. 


§  III.]  PRINCIPLES  OF  IMPULSE  AND  MOMENTUM.  337 

Let  v'  and  v"  denote  values  of  the  velocity  at  times  t[  and  t'[ 
respectively,  then 

jr/Vdt=mv/'-mv/ (i) 

The  integral  is  the  x  component  of  the  impulse  of  the  resultant 
for  the  interval  (/"  —  /'), and  the  right-handmember  is  the  incre- 
ment in  the  momentum  of  the  particle  along  the  x  axis  during 
that  interval.  Now  the  component  along  any  line  of  the  im- 
pulse of  the  resultant  of  any  number  of  forces  applied  to  a  par- 
ticle equals  the  algebraic  sum  of  the  components  of  the  impulses 
of  the  forces  along  the  same  line  ;*  hence 

The  algebraic  sum  of  the  components  along  any 
line   of  the   impulses  of  the  forces  applied  to  a 
particle  equals  the  increment  in  the  component  of 
the  momentum  of  the  particle  along  that  line. 
From  art.  236  we  have  also 

Ry=mdvy/dt     and     Rz=mdvg/dt; 
hence  RyZ = mzdvy/dt,     R^y  =  mydvg/dt, 

and  (Rgy — RyZ)  =  (mydvz  —  mzdvy)/dt. 

Now  the  left-hand  member  is  the  moment  of  the  resultant  about 
the  X  axis,  and  we  will  replace  it  by  M^.  The  right-hand  member 
equals  d{mVgy  —  mVyZ)/dt,  i.e.,  the  time-rate  of  increase  of  the 
moment  of  momentum  about  the  x  axis  (see  art.  323),  and  for 
convenience  we  will  denote  this  moment  by  U.     Then 

M^  =  dUldt,     or     MJt  =  dU, 

and  j^''M^dt  =  U''-U' (2) 

The  left-hand  member  equals  the  moment  of  the  impulse  of  the 
'resultant  about  the  x  axis  (equals  also  the  sum  of  the  moments 
of  the  impulses  of  the  forces  acting  on  the  particle),  and  the 
right-hand  member  is  the  increment  in  the  moment  of  the  mo- 
mentum of  the  particle  about  the  x  axis.     Hence 

"^  '^he  proof  can  be  supplied  by  the  reader  (see  art.  310) 


33^  "   IMPULSE  AND  MOMENTUM.  [Chap.  XV, 

The  algebraic  sum  of  the  moments  of  the  impulses 
of  the  forces  applied  to  a  particle  about  any  line 
equals  the  increment  in  the  moment  of  the  mo- 
mentum of  the  particle  about  that  line. 

328.  Principles  for  a  System  of  Particles. — The  internal  forces 
in  a  system  of  particles  occur  in  pairs,  the  forces  of  each  being 
at  each  instant  equal,  opposite,  and  collinear.  Therefore  the 
impulses  of  each  pair  of  forces  for  any  interval  are  equal  and 
opposite;  also  the  moments  of  the  impulses  of  each  pair  about 
any  line  are  equal  and  opposite.  It  follows  that  the  internal 
forces  contribute  nothing  to  any  change  in  the  momentum  or 
moment  of  momentum  of  a  system  of  particles,  and  it  can  be 
readily  shown  from  the  results  reached  regarding  a  single  par- 
ticle that 

A.  For  any  period  the  algebraic  sum  of  the  components  of 
the  impulses  of  the  external  forces  acting  on  a  system  along  any 
line  equals  the  increment  in  the  component  of  the  momentum 
of  the  system  along  that  line. 

B.  For  any  period  the  sum  of  the  moments  of  the  impulses 
of  the  external  forces  acting  on  any  system  about  any  line 
equals  the  increment  in  the  moment  of  the  momentum  about 
that  line. 

Special  Case:  No  External  Forces  Applied  to  the  System. — 
It  follows  from  the  foregoing  that 

(a)  The  component  of  the  momentum  of  the  system  along 
any  line  remains  constant;  this  principle  is  called  that  of  "con- 
servation of  linear  momentum." 

(h)  The  moment  of  the  momentum  of  the  system  with  re- 
spect to  any  line  remains  constant ;  this  principle  is  called  that 
of  "conservation  of  angular  momentum." 

§  IV.     Applications. 

329.  Computation  of  Velocity  and  Time. — The  equations 
expressing  the  principles  of  impulse  and  momentum  (^4  and  i?, 
art.  328)  contain  impulse  (or  moment  of  impulse)  terms  on  one 
side  and  momentum  (or  moment  of  momentum)  terms  on  the 
other.     The  factors  in  the  impulse  or  moment  of  impulse  terms 


§IV.]  APPLICATIONS.  339 

are  time  and  force  or  moment  of  force,  and  those  in  the  momen- 
tum or  moment  of  momentum  terms  are  velocity  and  mass  or 
moment  of  mass.  If  all  the  factors  except  one  in  one  of  the 
equations  are  known,  that  one  may  be  computed.  In  the  fol- 
lowing examples  the  unknown  quantity  is  a  velocity  or  a  time, 
and  the^examples  can^be  most  readily  solved  by  the  principles  of 
impulse  and  momentum. 

EXAMPLES. 

1.  A  body  weighing  80  lbs.  is  moved  along  a  horizontal  sur- 
face by  a  horizontal  pull  of  100  lbs.  If  the  frictional  resistance 
is  20  lbs.,  how  much  velocity  does  the  body  acquire  in  10  seconds  ? 

Solution :  There  are  three  external  forces  acting  upon  the 
body,  the  pull,  the  weight,  and  the  reaction  of  the  surface.  The 
components  of  the  impulses  of  these  along  the  path  are  respect- 
ively 

100X10,     o,     and     —20X10  Ib.-secs. 

Calling  the  velocity  of  the  body  at  the  beginning  of  the  lo-sec. 
period  v^,  and  that  at  the  end  of  it  v^,  the  equation  of  impulse 
and  momentum  for  the  period  is 

1000  —  200  =  2.4  5Z'2  —  2.457/1, 

2.45  being  the  mass  of  the  body  in  geepounds.  Hence  the  in- 
crement of  the  velocity  is 

z^2— ■^1  =  800/2.45  =  3^6.5  ft.-per-sec. 

2.  Solve  ex.  i ,  supposing  that  v^  =  o,  and  that  the  pull  instead 
of  being  constant  equals  20  +  10^  {t  being  time  in  sees,  after  the 
instant  of  starting). 

3.  Supposing  that  v^  in  ex.  i  is  300  ft.-per-sec,  determine 
how  long  the   body  would   slide   after   the   tenth   second,  the  ^ 
motion  taking  place  under  the  influence  of  friction  alone.    Q  ''  -  ^ 

4.  Suppose  that  the  disk  of  ex.  2,  art.  298,  is  rotating  en  a 
shaft  4  in.  in  diameter,  and  that  the  axle  friction  is  30  lbs.  In 
how  many  seconds  would  the  frictional  resistance  stop  it  ?  (Use 
B,  art.  328.) 

5.  Two  pulleys  are  mounted  on  the  same  shaft,  one  being 


340  IMPULSE  ^ND  MOMENTUM.  [Chap.  XV. 

"  fast  "  and  the  other  "  loose."  Suppose  that  the  shaft  (and  the 
fast  pulley)  to  be  turning  at  a  certain  instant  with  an  angular 
velocity  oj,  and  that  the  loose  pulley  is  quickly  made  fast  so  that 
it  also  turns  with  the  shaft.  If  the  shaft  turns  in  smooth  bear- 
ings, determine  the  subsequent  angular  velocity  of  the  pulleys 
and  shaft.  • 

Solution:  Under  the  supposition  there  are  no  external  forces 
acting  on  the  rotating  system  having  moments  about  the  axis 
of  the  shaft.  Then  the  moment  of  momentum  of  the  three 
bodies  about  the  axis  of  the  shaft  remains  constant. 

Let  /  denote  the  moment  of  inertia  of  the  shaft  and  the  fast 
pulley,  r  that  of  the  loose  pulley,  and  CO2  the  final  velocity.  Be- 
fore and  after  the  loose  pulley  is  made  fast,  the  moments  of 
momentum  of  the  three  bodies  about  the  axis  are  respectively 

Ico^  +  o     and     1 0)2+ 1' (02', 
and  since  these  are  equal  as  above  explained, 
co2=IwJ{I-\-r). 

6.  Two  spheres  whose  masses  and  radii  equal  m  and  r  respect- 
ively rotate  on  a  light  frame  about  a  vertical  axis  as  shown  in 
fig.  243  with  an  angular  velocity  10^.     Suppose  that,  in  some  way 

K- a- — -♦  * »"—  J 


m 


m 


Pig.   243. 

without  interfering  directly  with  the  motion,  the  distances  a 
are  increased  to  h.  Determine  the  angular  velocity  after  the 
change. 

330.  Pressures  Due  to  Jets. — Pressure  due  to  a  jet  of  liquid 
can  often  be  most  readily  determined  by  the  principles  of  im- 
pulse and  momentum.  How  this  is  done  is  explained  in  the 
solution  of  some  of  the  following 

EXAMPLES. 

I.  Fig.  244(a)  represents  a  jet  impinging  on  a  flat  surface  so 
that  the  direction  of  the  jet  is  changed  90®.     If  W  is  the  weight 


IV.] 


APPLICATIONS. 


341 


(of  water)  impinging  per  unit  time  and  v  the  impinging  velocity, 
show  that  the  pressure  of  the  water  on  the  plate  is  Wv/g. 

Solution:  Consider  the  motion  of  the  amount  of  water  rep- 
resented in  the  figure  for  a  small  period  dt,  during  which  it  moves 


DJI^ 


^   X 


(a) 


into  the  position  indicated  by  the  dotted  lines.  Let  MJ  and 
Mx"  denote  the  initial  and  final  momentums  in  the  x  direction 
of  the  water  whose  motion  is  being  considered,  then 

Mx  =  the  *  'x  momentum  "  of  5  +  ( Wdt/g)v ; 
also  Mx"  =  the  '  'x  momentum' '  of  -B , 

because  C  and  D  have  no  x  momentum.  Hence  the  change  in 
the  X  momentum  of  the  body  of  water  is 

Mx'-Mx"  =  (Wdt/g)v. 

The  impulse  of  the  force  producing  this  change  (the  pressure  of 
the  surface  on  the  water,  which  call  F)  is  Fdt\  hence 

Fdt  =  (Wdt/g)v, 
or  F  =  Wv/g. 

2.  Fig.  244(6)  represents  a  jet  impinging  on  an  inclined  sur- 
face. If  W  is  the  weight  of  the  water  impinging  per  unit  timcj 
and  V  the  impinging  velocity,  show  that  the  normal  pressure  on 
the  jet  equals  (Wv/g)  cos  (j>. 

Solution:  Consider  the  motion  of  the  body  of  water  repre- 
sented for  a  small  period  during  which  it  moves  into  the  position 
indicated  by  the  dotted  lines.  With  notation  as  in  the  preced* 
ing  solution, 

Mx  =x  momentum  oiB  +  (Wdt/g)v  cos  (f) 
and  Mx"  =  x  momentum  of  B. 


342 


IMPULSE  AND  MOMENTUM. 


[Chap.  X^ 


Ji- 


oo' 


oo 


Hence  the  change  in  the  x  momentum  of  the  body  of  water  is 

M^'  - MJ'  =  (Wdt/g)v  cos  i>. 

CaUing  the  normal  pressure  of  the  surface  on  the  water  F^,  the 
impulse  of  Fr,  is  Fndt,  and 

Fndt  =  {Wdt/g)v  cos  (j>,     or     F„,  =  (Wv/g)  cos  ((>. 

3.  Fig.  244(c)  represents  a  jet  impinging  on  a  guide  or  vane 
which  suppose  smooth,  so  that  the  speed  of  the  water  is  not 
changed.  Determine  the  x  and  y  components  of  the  pressure 
of  the  jet.  Ans.  The  x  component  is  Wv{i  —cos  ^)/g. 

4.  Fig.  245  represents  in  plan  and  elevation  a  simple  reaction- 
wheel.  Water  is  poured  in  at  the  top 
and  escapes  through  two  orifices  O  and 
O  in  a  horizontal  direction.  Such  a 
wheel  is  caused  to  revolve  by  the  "re- 
action" of  the  jets.  It  is  required  to 
determine  the  moment  of  these  reactions. 

Solution :  Let  W  denote  the  weight 
of  the  water  escaping  and  entering  per 
unit  time,  and  v  the  velocity  of  the 
escaping  water  relative  to  the  orifices. 
Then  if  co  denotes  the  angular  velocity 
of  the  wheel,  the  absolute  velocity  of  the 
^'-'^^  jets  is  roj—v.     Consider  the  motion,  for 

Fig.  245.  a  short  period,  of  the  water  in  the  wheel 

and  the  amount  about  to  flow  in  for  that  period  (Wdt);  that 
water  is  represented  in  the  elevation  fig.  245.  At  the  end  of 
the  period  the  entering  water  is  all  in  the"  wheel  and  an  amount 
Wdt  has  escaped,  as  vshown  in  the  plan.  Let  Af  and  Af  denote 
the  moment  of  momentum  of  the  bod}^  of  water  being  consid- 
ered about  the  axis  at  the  beginning  and  end  of  the  period  di 
Then  supposing  that  the  water  enters  vertically  or  so  that  the 
iet  is  bisected  by  the  axis  of  rotation, 

M'  =  the  moment  of  momentum  of  B 
and      M"=  "         '•        "  "  ''B-{-(Wdt/g)(r(o-v)r; 

the  change  of  moment  of  momentum  hence  equals 
iWdt/g){rco-v)r. 


ELEVATION 


§  IV.]  APPLICATIONS.  343 

Let  Ma  denote  the  moment  (about  the  axis)  of  the  pressure  of 
the  wheel  on  the  water;  then  the  moment  of  the  impulse  of 
the  pressure  is  Madt  and 

Madt^{Wdt/g){roj-v)r, 
or  Ma-={W/g){rco-v)r, 

and  the  moment  of  the  *water-pressure  (turning  the  wheel)  is 

{W/g){v  —  roj)r. 

5.  Show  that  when  water  issues  from  an  orifxe  in  a  vessel  at 
rest,  the  water  exerts  on  the  vessel  a  force  equal  to  {W/g)v 
in  a  direction  opposite  to  that  of  the  velocity  of  the  jet.  {W  de- 
notes the  weight  of  water  escaping  per  unit  time,  and  v  its 
velocity.) 

331,  Sudden  Impulses. — The  impulse  of  a  force  which  acts 
for  a  very  short  time  is  called  a  sudden  impulse. 

If  a  body  is  subjected  to  a  blow  and  to  an  ordinary  or  steady 
force  at  the  same  time,  the  impulse  of  the  latter  during  the  blow 
is  often  negligible  compared  with  the  sudden  impulse.  Thus  in 
the  case  of  a  ball  thrown  against  a  wall  there  are  two  forces 
acting  on  the  former  during  the  impact,  namely,  the  weight  of 
the  ball  and  the  reaction  of  the  v/all.  Supposing  the  ball  to  be 
thrown  horizontally,  the  reaction  of  the  wall  is  horizontal,  and 
its  impulse  equals  the  change  in  the  horizontal  component  of 
the  momentum  of  the  ball.  The  impulse  of  the  weight  equals 
the  change  in  the  vertical  component  of  the  momentum.  Now 
this  latter  change,  as  we  see  from  observation,  is  practically  zero 
compared  to  the  first  change ;  hence  the  impulse  of  the  weight 
is  also  practically  zero  compared  to  the  impulse  of  the  blow. 

The  principles  of  impulse  and  momentum  are  especially 
adapted  to  questions  involving  sudden  changes  of  motion,  and 
the  remainder  of  this  chapter  relates  to  changes  of  this  kind. 

332.  Force  of  a  Blow  and  Recoil  of  a  Gun. — When  one  body 
strikes  another  we  name  the  act  "a  blow,"  and  by  force  of  the 
blow  we  mean  the  pressure  between  the  bodies  during  the  blow. 
This  pressure  is  variable,  changing  from  zero  to  a  maximum  and 


344  llAPULSE  AND  MOMENTUM.  [Chap.  XV. 

back  to  zero.     If  t  denotes  the  duration  of  the  blow  andF  the 

variable  value  of  the  force,  the  impulse  is   /  Fdt.     By  average 

force  of  the  blow  *  is  meant  a  constant  force  which,  acting  for 
a  time  equal  to  the  duration  of  the  blow,  has  an  impulse  equal 
to  that  of  the  actual  force.  Thus  if  Fa  denotes  the  average 
force  of  the  blow, 

Fat=jydt. 

When  a  gun  is  fired,  the  powder-gases  exert  a  backward  force 
on  the  gun  as  well  as  a  forward  one  on  the  shot.  If  the  mass  of 
the  powder  were  negligible,  the  two  forces  at  each  instant  and 
their  impulses  would  be  equal.  Thus  let  m^  and  m^  denote 
masses  of  the  shot  and  gun  respectively,  and  v^  and  v^  their 
velocities  just  after  the  shot  leaves  the  barrel.  Then,  since  the 
impulses  producing  the  velocities  are  equal,  the  momenta  of 
shot  and  gun  must  be  equal,  i.e., 

If  a  gun  is  suspended  in  a  horizontal  position  by  means  of 
two  parallel  cords  of  equal  length,  the  velocity  of  recoil  {v^)  and 
the  height  (h)  to  which  the  gun  rises  during  the  recoil  are  related 
thus:  v^  =  2gk  (see  ex.  9,  art.  312);   hence 

mjV^  =  m2\/2gh,     or     v^  =  {m2/ni^\/  2gh. 

EXAMPLES. 

1.  A  body  whose  weight  is  W  is  dropped  twice  from  a  height 
k,  once  striking  on  a  pile  of  hay  and  once  upon  the  ground.  If 
the  times  of  the  impacts  are  f  and  t" ,  compute  the  average  force 
of  the  blow  in  each  case. 

2.  If  a  2-oz.  lead  bullet  strikes  a  plate  with  a  velocity  of  1000 
ft.-per-sec,  and  is  "flattened  out"  in  i/ioo  sec,  what  is  the 
average  force  of  the  blow  ? 

333.  Collision  or  Impact. — Consider  two  bodies  in  a  collision 
such  that  the  only  sudden  impulses  involved  are  those  of  the 
pressures  which  they  exert  upon  each  other.  Then  the  impulse 
of  other  forces  (as  gravity)  which  may  act  on  the  bodies  during 

*  This  average,  it  should  b^  noted,  is  a  time-average,  and  the  average 
force  of  the  chapter  on  Work  and  Energy  (XIV)  is  a  space-average. 


S  IV.]  APPLICATIONS.  345 

collision  are  negligible  and  the  principles  of  conservation  apply. 
In  such  a  collision  there  is  no  change  in  the  momentum  or  mo- 
ment of  momentum  of  both  bodies  together. 

Definitions :  If  the  mass-centres  of  two  bodies  before  collision 
move  aloi.g  the  same  straight  line,  the  impact  is  called  direct. 
If  the  forms  of  bodies  are  such  that  the  pressures  which  they 
exert  upon  each  oth^r  are  directed  along  the  line  joining  their 
mass-centres,  the  impact  is  called  central. 

334.  Direct  Central  Impact. — Let  the  common  path  of  the 
mass-centres  be  taken  as  an  x  axis,  then  the  momentum  of  each 
body  has  no  y  ov  z  component.  Since  the  impulses  due  to  Ihe 
collision  are  directed  along  the  x  axis,  the  increment  in  the 
momentum  of  each  body  due  to  the  collision  is  directed  along 
the  X  axis  and  the  momenta  of  the  bodies  after  the  collision  have 
no  y  ov  z  components,  i.e.,  the  mass-centres  of  the  bodies  move 
along  the  x  axis  after  the  impact. 

The  moment  of  the  impulse  of  the  force  exerted  upon  either 
body  about  any  axis  through  its  mass-centre  is  zero ;  hence  the 
moment  of  momentum  of  each  body  about  any  axis  through  its 
mass-centre  is  unchanged  in  the  collision.  In  particular,  if  the 
motion  of  either  body  before  collision  is  translatory  it  will  be 
so  after  collision. 

Let  A  and  B  be  two  translating  bodies  in  collision  and  let 
m^  and  Wj  denote  the  masses  of  A  and  B ; 
t'l  and  i/j    #     *'       **  velocities  of  ^  and  5  before  impact; 
v^'  and  V2         *'       "  **         "  A  and  B  after  impact. 

The  velocities  should  be  regarded  as  having  sign,  those  in  one 
direction  being  positive  and  those  in  the  other  negative.  As 
previously  explained,  the  total  momentum  of  the  two  bodies  is 
not  changed  by  the  impact ;  hence 

m{i\  +  m2V2  =  m{Vi  +m2V2^ (i) 

It  has  been  determined  experimentally  that  when  two  spheres 
collide  directly  and  centrally  the  velocity  of  either  relative  to 
the  other  is  reversed  by  the  impact  and  diminished,  and  that  the 
diminution  depends  only  on  the  materials.     That  is, 

(i\-v^)^-e{v,'-v./), (2) 

e  being  a  common  fraction  and  called  "coefficient  of  restitution." 


346  IMPULSE  AND  MOMENTUM.  [Chap.  XV. 

We  assume  that  relative  velocities  of  any  two  bodies  before  and 
after  a  direct  and  central  impact  are  related  as  in  the  case  of 
spheres.  * 

Equations  (i)  and  (2)  determine  the  final  velocities  of  two  col- 
liding bodies  in  terms  of  their  masses,  initial  velocities,  and  their 
coefficient  of  restitution.     Thus  we  find  from  the  equations  that 

V='^l-(l  +^)(^l -'^2)^2/(^1 +^2)»        •        •        •        (3) 

v^'  =  v^-{i+e){y^-v^)mjim^+m^).     ...     (4) 

The  value  of  the  impulse  equals  the  change  in  the  momentum 
of  either  body.     This  change  is 

m{u^'-m^v^=  -(i  -\-e){Vy^-v^)m^mJ{m^-^m^).       .     (5) 

During  a  collision  each  body  is  first  compressed,  and  after 
full  compression  has  been  reached  it  begins  to  recover  its  natu- 
ral form  unless  the  body  is  perfectly  inelastic.  The  time  occu- 
pied by  the  compression  is  called  the  period  of  compression,  and 
it  is  assumed  to  be  alike  for  both  bodies.  The  remainder  of  the 
time  of  an  impact  is  called  the  period  of  restitution.  When  the 
compression  ends  and  restitution  begins  the  velocities  of  the 
mass-centres  are  the  same;  let  v  denote  this  common  velocity. 
Then,  from  the  principle  of  conservation, 

hence  7;=-^-^- — — =  — '-^ (6) 

For  the  case  of  a  body  (mj  impinging  on  a  fixed  one  (Wj), 
we  substitute  for  Wj  and  V2  respectively  00  and  o;  then 

v^'=-ev^ (7) 

335.  Loss  of  Energy  in  an  Impact. — There  is  always  a  loss  of 
kinetic  energy  in  an  impact  (unless  e  =  i),  its  value  being  found 
as  follows:    The  kinetic  energy  before  and  after  impact  equals 

^m{U{^  +  iw2^'2^     and     ^n^u^  ^  +  hm^v^  ^. 

Subtracting  the  latter  from  the  former  and  substituting  for  v^ 
and  V2  their  values  from  (3)  and  (4),  we  have  as  the  loss 


§iv.J 


APPLICATIONS. 


347 


^ 


EXAMPLES.* 

I .  Show  that  if  two  bodies  of  equal  mass  and  perfectly  elastic 


(^  =  i)  collide,  they  exchange  velocities. 

2.  If  a  ball  falls  from  a  height  h  upon  a  horizontal  plane, 
show  that  the  height  of  rebound  equals  e^h. 

3.  Two  bodies  of  unequal  mass  with  momenta  numerically 
equal  meet.  Show  that  their  momenta  after  impact  are  still 
numerically  equal. 

I  t  4.  A  body  whose  mass  is  10  lbs.,  moving  with  a  velocity  of 
''  87  ft.-per-sec,  overtakes  a  body  whose  mass  is  40  lbs.,  moving 
with  a  velocity  of  6  ft.-per-second.  The  coefficient  of  restitu- 
tion being  3/4,  compute  the  final  velocity  of  each  and  the  im- 
pulse of  the  forces  due  to  impact.  Ans.  v^'  =  6.^]  ft. /sec. 
|.'7  5.  Suppose  that  in  ex.  4  the  bodies  meet  and  collide;'  then 
solve. 

6.  Suppose  that  in  ex.  4  the  bodies  are  inelastic,  and  solve. 

7.  Deduce  values  of  the  impulses  for  the  periods  of  com- 
pression and  restitution,  and  show  that  the  former  is  independ- 
ent of  the  coefficient  of  restitution. 

336.  Ballistic  Pendulum  and  Centre  of  Percussion. — Fig. 
246  represents  a  ballistic  pendulum  (for  determining  the  ve- 
locity of  a  projectile);  let  M  denote  its  mass,  k  its 
radius  of  gyration  with  respect  to  its  axis  of  rota- 
tion, m  the  mass  of  the  shot,  v  its  striking  velocity, 
and  oj  the  angular  velocity  produced  in  the  pendu- 
lum by  the  impact.  The  velocity  of  the  shot  just 
after  it  is  imbedded  and  has  come  to  rest  relative 
to  the  pendulum  equals  rco.  Since  the  time  of  the 
impact  (during  which  the  angular  velocity  is  gen- 
erated) is  very  short,  the  angular  displacement  of 
the  pendulum  during  that  time  is  practically  zero,  ^'^^-  ^46. 
and  the  direction  of  the  velocity  of  the  shot  just  after  imbedding 
is  practically  horizontal.  Hence  the  impulse  exerted  on  the 
shot  by  the  pendulum  and  that  exerted  on  the  pendulum  by  th< 
shot  equal 

m{v  —  r(tj). 


F 


J- 


*  Direct  central  impact  is  implied  in  all  the  examples. 


348  -  IMPULSE  AND  MOMENTUM.  [Chap.  XV. 

The  moment  of  the  impulse  of  the  shot  on  the  pendulum  with 
respect  to  the  axis  is  m(v  —  roj)r,  and  the  change  in  the  moment 
of  momentum  of  the  pendulum  (witR  respect  to  the  axis)  dur- 
ing the  time  of  the  impulse  equals  Mk^w  (see  art.  326).  Ac- 
cording to  B,  art.  328, 

m{v  —  roj)r  =  Mk'^oj,     or     v  =  {Mk^+mr^)cu/mr. 

If  h  is  the  height  to  which  the  centre  of  gravity  of  the  penduluHi 
rises,  co=^{i/k)V2gh  (see  ex.  8,  art.  312);   hence 

Mk^+mr''   /— , 

V= r— V  2m, 

mrk 

and  from  this  equation  v  may  be  computed,  the  quantities  on 
the  right  side  being  readily  measured  in  any  actual  c  se. 

Centre  of  Percussion . — Let  R^  and  R^'  denote  the  average 
values  of  the  vertical  and  horizontal  components  f  the  hinge 
reaction  on  the  ballistic  pendulum  during  an  impcxt,  P  the  aver- 
age value  of  the  force  of  the  blow,  and  t  the  duration  of  the  blow 
which  is  assumed  to  be  so  short  that  the  displacement  of  the 
pendulum  for  the  time  t  is  practically  zero. 

The  momentum  of  the  pendulum  after  the  blow  is  Mrco  and 
its  direction  is  horizontal,  and  the  moment  of  its  momentum 
about  the  axis  is  Mk^oj  (see  art.  326);  hence,  according  to  the 
principles  of  impulse  and  momentum  (see  fig.  246), 

R't-Wt  =  o, 
Pt-R"t==Mroj, 
(Pt)r  =  Mk'oj. 

From  these,  R'  =  W  and  R"  =  Moj{kyr-r)/U 

Observe  that  R'  is  independent  of  the  blow  and  that  i?" 
-equals  zero  if  the  blow  is  applied  at  a  distance  equal  to  k^/r 
below  the  axis.  The  point  in  OC  produced,  whose  distance 
from  0  equals  k"^/?,  is  called  centre  of  percussion;  it  coincides  with 
the  centre  of  oscillation  of  the  pendulum  (see  art.  267),  and  the 
methods  given  in  art.  267  may  be  employed  to  locate  the  centre 
of  percussion  of  a  ballistic  pendulum. 


APPENDIX  A. 

VECTORS. 

A  I.  Scalar  and  Vector  Quantities. — A  quantity  which  has  magni- 
tude or  magnitude  and  sign  only  is  called  a  scalar  quantity.  An 
amount  of  money,  a  volume,  etc.,  are  examples  of  scalar  quantities. 

A  quantity  which  has  magnitude  and  direction  is  called  a  vector 
quantity.  A  step,  a  force,  and  a  velocity  are  examples  of  vector 
quantities. 

The  methods  of  ordinary  algebra  are  sufficient  for  purposes  of 
analysis  in  mechanics  when  only  scalar  quantities  are  concerned, 
but  insufficient  in  general  for  dealing  with  vector  quantities.  For 
example,  the  algebraic  sum  of  two  forces  is  in  general  meaningles? 
or  at  any  rate  without  mechanical  significance,  but  their  vectorial 
sum  (to  be  explained)  has  a  very  important  significance.  There  is 
a  branch  of  mathematics  sometimes  called  Vector  Algebra,  the 
methods  of  which  are  especially  adapted  for  dealing  with  vector 
quantities.  We  proceed  to  a  brief  explanation  of  Addition  and 
Subtraction  by  those  methods. 

A  2.  Vector  Defined. — A  straight  line  of  definite  length  and 
direction  is  called  a  vector.  The  word  direction  here  refers  not 
only  to  the  inclination  (or  "clinure")  of  the  line,  but  also  to  its 


(a)  (b) 

Fig.  A  i. 

"  ('*  right-  or  Isft-ness  "  or  **  up-  or  down-ness  "  along  the  line). 
The  sense  is  usually  indicated  by  an  arrow-head  placed  on  the  line. 
The  lines  of  fig.  A  i  are  vectors. 

Two  vectors  in  order  to  be  equal  must  be  equal  in  length  and 
of  the  same  direction.  The  first  and  second  vectors  of  fig.  A  i  (a) 
(counting  downwards)  are  equal  because  they  agree  in  length  and 

349 


350 


/iPPENDIX  A. 


direction,  but  the  third  is  not  equal  to  either  of  the  others  because 
its  direction  is  different  from  theirs. 

A  3.  Addition  of  Vectors. — Definition. — The  sum  of  the  vectors 
AB  and  BC  is  the  vector  AC  (fig.  A  i  6).  Notice  that  this  defini- 
tion does  not  conflict  (as  at  first  sight  it  may  appear)  with  the 
proposition  of  geometry  which  states  that  the  length  of  any  side 
of  a  triangle  is  less  than  the  sum  of  the  lengths  of  the  other  two. 

To  add  more  than  two  vectors  we  proceed  as  in  the  ordinary 
algebra,  i.e.,  we  add  any  two,  then  to  that  sum  another  vector,  and 
so  on  until  all  have  been  combined.  As  in  the  addition  of  scalars, 
the  sum  of  several  vectors  does  not  depend  on  the  order  in  which 
they  are  added;  thus  the  sum  of  the  four  vectors  a,  6,  c,  and  d  (fig. 
A  2)  is  found  to  be  AB  by  adding  them  in  a  certain  order  and  CD 


Fig.  a  2. 


by  adding  in  another  order,  and  AB  and  CD  are  found  to  agree  in 
length  and  direction.  If  they  are  added  in  any  other  order,  the 
sum  will  be  found  to  be  equal  \o  AB  or  CD. 

EXAMPLE. 

Reverse  the  arrows  on  vectors  c  and  d  (fig.  A  2)  and  add  the 
four  vectors  in  at  least  two  ways. 

A  4.  Negative  of  a  Vector. — In  ordinary  algebra,  the  negative  of 
a  quantity  is  one  which  added  to  the  quantity  gives  a  sum  equal 
to  zero.  So,  too,  the  negative  of  a  vector  is  one  which  added  to 
the  vector  gives  a  sum  equal  to  zero.  Now  the  sum  of  two  vectors 
which  are  equal  in  length,  parallel,  and  opposite  in  sense  equals 
zero ;  hence  either  is  the  negative  of  the  other. 

A  5.  Subtraction  of  Vectors. — In  ordinary  algebra,  a  quantity  is 
subtracted  from  another  by  adding  its  negative.  So,  too,  to  sub- 
tract a  vector  from  another  we  add  the  negative  of  the  former  to 
the  latter. 

EXAMPLES. 

1.  Subtract  vector  a  (fig.  A  2)  from  vector  6. 

2.  Subtract  vector  h  from  vector  a. 


APPENDIX  B. 

RATES.* 

The  term  rate  is  in  common  use,  but  usually  m  an  inexact  sense 
In  the  following  a  precise  meaning  is  given  to  it  which,  it  will  be 
observed,  does  not  conflict  with  the  -popvlax  notions  in  so  far  as 
they  are  exact. 

B  I.  Kinds  of  Variable  Quantities. — Let  X  and  y  be  two  quanti- 
ties which  are  related  to  each  other,  a  change  in  x  producing  a 
change  in  y.  If  all  equal  changes  in  x  (large  or  small)  produce 
equal  changes  in  y,  y  is  said  to  vary  uniformly  with  respect  to  x 
and  it  is  called  a  uniform -variable.  If  equal  changes  in  x  produce 
unequal  changes  in  ;v,  it  is  said  to  vary  non-uniformly  with  respect 
to  x,  and  it  is  called  a  non-uniform  variable. 

B  2.  Rate  of  a  Uniform  Scalar. — If  y  is  a  uniform  variable,  then 
the  locus  representing  the  relation  between  x  and  ^  is  a  straight 
line,  for  in  such  a  locus  equal  changes  in  x 
produce  equal  changes  in  y. 

In  this  simple  case,  the  popular  meaning 
of  "the  rate  of  y"  is  definite,  it  being  the 
change  in  y  per  unit  change  in  x.  If  Jx, 
x^—Xi  (see  fig.  B  i),  denotes  any  change  in  x, 
and  Jy,  y2—yi,  the  corresponding  change  in 
y,  then  the  change  in  y  per  unit  change  in  x 
is  'dy/Jx,  or  (y2—yi)/{x2—x^).  Hence  if  r 
denotes  rate, 

T  =  Jj/Jx (i) 

Evidently  r  is  the  same  for  all  values  of  ^x,  i.e.,  the  rate  of  a  uni- 
form variable  is  constant. 

B  3.  Rate  of  a  Non- Uniform  Scalar. — If  y  denotes  a  non-uniform 
variable,  then  the  locus  representing  the  relation  between  x  and  y 
obviously  must  be  curved. 

According  to  popular  notions,  the  rate  of  a  non-uniform  vari- 

*  This  appendix  is  intended  for  students  who  do  not  associate  the 
idea  of  rate  with  dy/dx,  but  do  understand  that  dy/dx  is  a  "limit." 

35^- 


*-Xr^ 


-<Cr — 


Fig.  B  I. 


352 


APPENDIX  B. 


able  is  not  constant  and  has  a  definite  value  at  each  value  of  the 
variable,  but  these  definite  values  of  the  rate  are  not  clearly  dis- 
cerned in  the  popular  mind. 

Average  Rate  Defined.— Let  Ay  denote  the  change  in  y  due  to  a 
change  4oc  in  x  (see  fig.  B  2  a).  Evidently  the  rate  of  a  uniform  vari- 
able, also  depending  on  x,  might  be 
such  that  the  change  in  that  variable 
due  to  a  change  Ax  in  x  equals  Ay. 
The  average  rate  of  the  non-uniform 
variable  for  the  range  X2  —x^  is  de- 
fined as  the  rate  of  such  a  uniform 
variable  as  just  mentioned.  Now 
the  rate  of  the  uniform  variable  is 
Ay /Ax;  hence  if  ra  denotes  average 
rate  of  the  non-uniform  variable, 

To 


/y^^y 

1/2 

^^^N^ 

y^A      ~Ax 

X- 

* — —- «s — -* 

(a) 

In 


— iCr— > 


■OCz- 


ib) 


Fig.  B  2. 


Ay/Ax.    ...      (2) 

True  or  Instantaneous  Rate  De- 
■fined. — The  value  of  the  average  rate 
of  a  variable  is  generally  different  for 
different  values  of  x^—x^,  but  (see 
the  figure)  the  average  rate.  Ay /Ax,  approaches  a  definite  value  as  Ax 
approaches  zero  {x^  approaches  x^,  for  instance).  The  rate  of  the 
variable  y  at  the  value  y  =  yx  is  defined  as  the  limit  of  the  average 
rate  of  y  for  a  range  x^  —x^  as  x^  approaches  x^.  Now  the  limit  of 
Ay /Ax  is  dy/dx,  i.e.,  the  rate  of  y  at  the  value  y  =  yi  equals  the  value 
of  dy/dx  corresponding  to  y  =  y^  (and  x  =  x^).  Hence  if  r  denotes  the 
rate  of  y  at  any  value  of  y, 

r  =  dy/dx (3) 

Observe  that  the  rate  of  y  at  the  value  y=yi  is  represented  by 
the  slope  of  the  tangent  to  the  locus  representing  the  relation  be- 
tween X  and  y  at  the  point  Xiyi. 

Consistency  of  the  Definitions. — The  average  value  of  a  number 
of  quantities  is  computed  by  adding  them  and  dividing  the  sum  by 
their  number — the  quotient  being  the  average  sought.  It  can  be 
shown  that  the  above  definitions  of  average  and  instantaneous  rates 
agree  with  this  method  of  computing  averages.  Thus  let  r^  and  r^ 
denote  the  values  of  the  rate  of  y  when  x^x^  and  X2  respectively 
(see  fig.  B  2  6),  and  let  ordinates  to  the  curve  represent  values  of 
the  rate  at  intermediate  values  of  x.  Then  since  the  average  ordi- 
nate represents  the  average  rate,  the  latter  is  given  by 
ra  =  (area  ABB.A^)  /{x.  -x^. 


RATES. 


353 


It  is  shown  in  w 
hence 


orks  on  calciilus  that  the  area  ABB^A^  equals  /   ^rdx, 


y\         X2  —Xi 


Ay 
Ix 


Thus  we  find  that  th    ordinary  method  of  computing  averages  leads 
to  a  r  -^ult  identical  with  tha    given  by  eq.  (2). 

The  definition  of  "  instantaneous  rate  "  further  agrees  with  popu- 
lar notions,  as  may  be  explained  thus:  Let  y  and  z  denote  uniform 
and  non-uniform  variables  respectively,  both  depending  on  x,  and 
suppose  that  their  relations  to  x  are  represented  by  the  straight  and 
curved  loci  of  fig.  B  3.  From  the  figure  it  is  plain  that  for  any 
change  Ax  between  A  and  B  (at  -4,  the  tan- 
gent to  the  curve  is  parallel  to  the  straight 
line)  the  change,  or  increment,  in  z  is 
smaller  than  that  in  y,  and  that  they  be- 
come more  nearly  equal  the  nearer  B  is 
to  A.  Also  for  any  change  Ax  between  A 
and  C  the  change  in  z  is  greater  than  that 
in  y,  and  they  become  more  nearly  equal 
the  nearer  C  is  to  ^.  Popularly  stated :  up 
to  A ,  the  rate  of  z  is  less  than  the  rate  of  y, 
beyond  A  the  rate  of  z  is  greater  than  that 
of  y,  and  at  A  they  are  equal.  Now  our  formulas  for  rate  should 
agree  with  this  popular  expression,  and  it  is  readily  shown  that 
they  do.  Thus  the  rates  of  0  and  ^  are  fi2r/(i^  and  J^z/J^;  and  it  will 
be  seen  from  the  figure  that  for  values  of  x  less  than  x'  the  former 
is  the  lesser,  for  values  of  x  greater  than  x'  the  former  is  the  greater, 
and  for  x  =  x'  they  are  equal. 

B  4.  Sign  of  a  Rate. — Both  Ay /Ax  and  dy/dx  may  be  negative  as 
well  as  positive;  hence  in  order  that  equations  (i),  (2),  and  (3)  may 
be  true  as  to  sign,  it  is  necessary  to  regard  a  rate  as  having  sign 
and  the  sign  must  be  the  same  as  that  of  the  expression  for  the  rate 
{Ay /Ax  or  dy/dx).  Now  Ay / Ax  (for  a  uniform  variable)  and  dy/dx 
are  positive  when  y  increases  as  x  increases,  and  they  are  negative 
when  y  decreases  as  x  increases.* 

Hence  the  rate  of  y  with  respect  to  x  at  any  particular 
value  of  y  is  positive  or  negative  according  as  y  in- 
creases or  decreases  at  that  value  as  x  increases. 


*  Algebraic  increase  and  decrease  are  meant. 


354  "  APPENDIX  B. 

Thus  in  fig.  B  i  (a)  the  rate  of  y  is  positive,  in  fig.  B  i  (6)  it  is  neg- 
ative, and  in  fig.  B  2  (a)  the  rate  has  different  signs  at  different 
values  of  rj;  or  3;,  it  being  positive  from  mto  n,  negative  from  n  to  o. 
B  5.  Unit  of  Rates. — The  expressions  for  rates  in  eqs.  (i),  (2), 
and  (3)  imply  a  certain  unit,  namely,  the  rate  of  a  uniform  variable 
y  which  changes  a  unit  in  amount  for  a  unit  change  in  x.  Thus  if 
y  denotes  volume  and  x  distance,  the  unit  rate  might  be  one  cubic 
foot  per  foot,  one  gallon  per  inch,  etc. ;  if  y  denotes  distance  and  x 
time,  the  unit  rate  might  be  one  foot  per  second,  one  mile  per  hour, 
etc.,  etc. 

B  6.  Rate  of  a  Uniform  Vector. — Let  y  denote  a  vector  which  is 
related  to  x  and  let  Oa,  Oh,  Oc,  etc.  (fig.  B  4),  represent  y  at  values  of 

X  equal  to  x^,  x.^,  x^,  etc.,  the  dif- 
ferences between  successive  values 
of  X  being  equal  (riTj  ""^i^^s  ~^2' 
etc.).  Now  if  ab=bc  =  cd  etc., 
y  is  a  uniform  variable,  for  the 
changes  (or  increments)  in  y  (vec- 
tors) for  equal  changes  in  x  are 
equal. 

-piQ    B  .  By  rate  of  a  uniform  vector  y 

is  meant  its  change  per  unit  change 
in  x;  hence  the  rate  equals  any  change  in  y  (a  vector)  divided  by 
the  corresponding  change  in  x,  as  (vector  ac) /{x^  —^1).  If  OA  and 
OB  represent  y  for  any  values  oi  x  as  x  and  x  +  Jx  respectively,  then 
the  rate  oi  y  is 

(vector  AB)/Jx. 

Let  P  be  a  fixed  point  in  ad  in  the  direction  BA  from  A ,  and  let 
5  denote  the  variable  distance  PA.  The  increment  of  5  due  to  a 
change  Jx  in  x  equals  AB  as  shown,  and  since  y  is  a  uniform  vari- 
able, ^  evidently  is  also;  hence  the  rate  of  5  is  As/ Ax.  Finally, 
since  As  =  length  AB,  the  rate  of  y  is  a  vector 

whose  direction  ic  that  of  the  increment  of  y  and 
whose  magnitude  equals  ds/dx. 

B  7.  Rate  of  a  Non- Uniform  Vector. — Let  Oa,  Oh,  Oc,  etc.  (fig.  B  5), 
represent  a  vector  ;y  at  values  of  x  equal  to  x^,  x^^,  x^,  etc.,  the 
successive  differences  in  x  being  equal.  The  changes  in  y  corre- 
sponding to  changes  x^—x^,  x^—x^,  etc.,  are  the  vectors  ah,  he,  etc. 
Then  if,  in  fig.  B  5  (a),  ah,  he,  etc.,  are  unequal  and  if,  in  Fig.  B  5  (6), 
ah,  he,  etc.,  are  equal  or  unequal,  >'  is  a  non-uniform  variable;   for 


RATES.  355 

the  changes,  or  increments,  in  y  due  to  equal  changes  in  x  are  not 
equal.* 

Average  Rate  Defined. — Jhe  average  rate  of  a  vector  y  for  any 
range  dx  in  x  is  defined  as  the  rate  of  a  uniform  vector  whose  change, 
for  the  same  range  in  x,  equals  that  of  y.     Thus  if  OA  and  OB  repre- 


sent y  for  values  x  and  x  +  Jx,  the  change  in  y  for  the  range  dx 
is  the  vector  AB.  Now  the  rate  of  a  uniform  vector  whose  change 
is  also  AB  for  the  range  Jx  is  (vector  AB)/Jx,  and  this,  by  the 
definition,  is  the  average  rate  of  y  during  the  change  Jx,  i.e., 

the  average  rate  is  a  vector  whose  direction  is  AB  and 

whose  magnitude  equals  (length  AB)  /  J x. 

Actual  or  Instantaneous  Rate  Defined. — The  value  of  the  average 

rate  of  y  is  different  in  magnitude  and  in  direction  for  different 

values  of  Jx.    The  average  rate  approaches  a  definite  direction  and 

a   definite   magnitude   as    Jx  approaches   zero.     The  rate   of  the 

variable  y  at  the  value  y  =  OA  is  defined  as  the  limit  of  the  average 

rate  of  3/  as  Jx  approaches  zero  (B  approaches  A).     The  limiting 

direction  of  the  average  rate  is  the  limiting  direction  oi  AB,  which 

is  the  direction  of  the  tangent  at  A.     The  limiting  value  of  the 

average  rate  is  the  limit  of  (chord  AB)/Jx,  which  is  the  same  as 

the  limit  of  (arc  AB)/Jx.     Now  let  P  be  any  fixed  point  on  the 

curve  AB  in  the  direction  BA  from  A,  and  5  the  distance  from  P 

to  A ;   then  the  change  in  s,  Js,  due  to  a  change  Jx  in  x  is  the  arc 

AB,  and 

,.      arc  AB     ,.     -.  Js     ds 

lim =  limit  —  =  -r-- 

JX  Jx     dx 

Finally,  the  rate  of  y  at  the  value  y  =  OA  is  a  vector 

whose  direction  is  that  of  the  tangent  at  A ,  and 
whose  magnitude  =ds/dx. 

*  Equal  vectors  are  equal  in  length  and  the  same  in  direction. 


356  APPENDIX  B. 

It  is  implied  that  the  fixed  point  from  which  5  is  measured  is 
so  taken  that  Js  is  positive.  The  arrow  on  the  tangent,  giving  the 
sense  of  the  rate,  points  in  the  positive  §  direction. 

B  8.  Descriptive  Terms. — A  variable  y  may  depend  on  several 
quantities.  If  so,  it  has  as  many  rates,  and  we  say  for  brevity  the 
"  X  rate  oiy"  to  distinguish  the  rate  of  y  with  respect  to  x  from  its 
other  rates.  The  rate  dy/dx  is  often  described  by  naming  the  kind 
of  a  quantity  represented  by  x\  thus  if  x  denotes  distance,  dy/dx  is 
called  a  "space  rate"  ot  y\  ii  x  denotes  time,  dy/dx  is  called  a 
"  time  rate  "of  y.  The  time  rate  of  y  is  sometimes  indicated 
thus,  y. 

If  y  and  x  denote  certain  quantities,  the  rate  is  given  a  single 
name;  thus  if  y  and  x  denote  distance  and  time  respectively,  dy/dx 
is  called  velocity;  if  y  denotes  mass  and  x  volume,  dy/dx  is  called 
density. 


APPENDIX  C. 

DIMENSIONS  OF  UNITS. 

C  I.  Magnitude  of  a  Quantity. — The  magnitude  of  a  quantity  is 
expressed  by  stating  how  many  times  larger  it  is  than  a  standard 
quantity  of  the  same  kind  and  naming  the  standard.  Thus,  we 
say  that  a  certain  distance  is"  lo  miles,  meaning  that  the  distance  is 
lo  times  as  great  as  the  standard  distance,  the  mile. 

The  number  expressing  the  relation  between  the  magnitude  of 
the  quantity  and  the  standard  (the  number  lo  in  the  illustration) 
is  called  the  numeric  (or  numerical  value)  of  the  quantity,  and  the 
standard  is  called  the  unit. 

C  2.  Fundamental  and  Derived  Units. — A  unit  for  measuring  any 
kind  of  quantity  may  be  selected  arbitrarily,  but  it  must  of  course 
be  a  quantity  of  the  same  kind  as  the  quantity  to  be  measured 
(e.g.,  a  unit  for  measuring  lengths  must  be  a  length).  Thus,  as 
unit  of  velocity  we  might  select  the  velocity  of  light,  as  unit  of  area 
the  area  of  one  face  of  a  silver  dollar,  etc.  Many  units  in  use  are 
arbitrarily  chosen,  i.e.,  without  reference  to  another  unit  (e.g.,  the 
bushel  and  the  degree),  but  it  is  convenient  practically  to  define 
them  with  reference  to  each  other.  All  mechanical  and  nearly  all 
physical  quantities  can  be  defined  in  terms  of  three  arbitrarily 
selected  units,  i.e.,  ones  not  dependent  on  any  other  units.  These 
are  called  fundamental  units,  and  the  others,  defined  with  reference 
to  them,  derived  units.  It  is  customary  in  works  on  theoretical 
mechanics  and  physics  to  choose  as  fundamental  the  units  of 

length,  mass,  and  time, 

but  it  is  sometimes  more  convenient  to  take  as  fundamental  the 
units  of 

length,  force,  and  time. 

In  the  following  article  we  give  a  discussion  of  derived  units  with 
reference  to  each  of  these  sets  of  fundamentals,  and  on  pages  360  and 
361  there  appear  summaries  in  which  the  absolute  units  are  re- 
ferred to  the  first  set  of  fundamentals  and  the  gravitational  units 
to  the  second  set.  But  either  set  might  serve  as  fundamentals  for 
all  absolute  and  gravitational  units. 

357 


ZS^  APPENDIX  C. 

C  3.  "  Dimensions  "  of  Units. — A  statement  of  the  way  in  which 
a  dsrived  unit  depends  on  the  fundamental  ones  is  called  a  state- 
ment of  its  dimensions. 

Obviously  an  area  depends  only  on  the  tinit  of  length,  definitely 
as  the  square  of  the  unit  of  length.     Thus 

(one  sq.  yd.) /(one  &q.  ft.)  =(one  yd.  or  three  ft.) '/(one  ft.) '  =  9. 

This  relation  is  expressed  in  the  form  of  a  "  dimensional "  equation, 

thus 

(unit  area)  =  (unit  length) ', 

and  briefly  a  unit  area  is  said  to  be  "two  dimensions  in  length." 
Similarly  a  unit  volume  is  said  to  be  three  dimensions  in  length. 

Velocity. — According  to  the  definition  of  velocity  (art.  167),  a 
unit  velocity  is  directly  proportional  to  the  unit  length  and  in- 
versely to  the  unit  time;  hence  if  V,  L,  end  T  denote  units  of  veloc- 
ity, length,  and  time  respectively,  the  dimensional  equation  is 

V=L/T=LT-S 
and  a  unit  velocity  is  one  dimension  in  length  and  minus  one  in  time. 

Acceleration. — According  to  the  definition  of  acceleration  (art. 
173),  a  unit  acceleration  is  proportional  directly  to  the  unit  veloc- 
ity and  inversely  to  the  unit  of  time ;  hence  if  A  denotes  unit  accel- 
eration, the  dimensional  equation  is 

A  =  V/T=L/T2=LT-», 

and  a  unit  acceleration  is  one  dimension  in  length  and  minus  two 
in  time. 

Angular  Velocity. — According  to  the  definition  of  angular  veloc- 
ity (art.  210),  a  unit  angular  velocity  is  proportional  directly  to 
the  unit  of  angle  and  inversely  to  the  unit  of  time ;  hence  if  (o  and  0 
denote  units  of  angular  velocity  and  angle  respectively,  the  dimen- 
sional equation  is 

(o  =  6/T,     or     (o  =  T-\ 

since  units  of  angle,  (degree,  radian,  etc.)  are  independent  of  the 
fundamental  units.  A  unit  angular  velocity  is  therefore  minus  one 
dimension  in  time. 

Angular  Acceleration. — According  to  the  definition  of  angular 
velocity  (art.  213),  a  unit  angular  acceleration  is  proportional 
directly  to  the  unit  angular  velocity  and  inversely  to  the  unit  time; 
hence  if  a  denotes  unit  angular  acceleration,  the  dimensional  equa- 
ion  is 

a  =  w/T  =  T-', 

and  a  unit  angular  acceleration  is  minus  two  dimensions  in  time. 


DIMENSIONS  OF  UNITS.  359 

Force.- — In  accordance  with  the  equation  of  motion  of  a  particle 
(art.  236),  R=ma,  or 

"  force  =  mass  X  acceleration  ", 

i.e.,  the  unitforce  is  directly  proportional  to  the  units  of  mass  and 
acceleration;  hence  if  F  and  M  denote  units  of  force  and  mass 
respectively,  the  dimensional  equation  is 

F=MA  =  LMT-', 

and  a  unit  force  is  one  dimension  in  length,  one  in  mass,  and  minus 
two  in  time. 

Mass. — If  we  regard  length,  force,  and  time  as  fundamental 
units,  then  the  last  equation  written  as  follows  is  the  dimensional 
equation  for  a  unit  mass : 

M=FT2A  =  L-^FT^ 

and  a  unit  mass  is  minus  one  dimension  in  length,  one  in  force,  and 
two  in  time. 

Work. — According  to  the  definition  of  work  (art.  2^6),  the  unit 
of  work  is  directly  proportional  to  the  units  of  force  and  length; 
hence  if  W  denotes  unit  work,  the  dimensional  equation  is 

W=LF=L2MT-2, 

and  a  unit  work  is  one  dimension  in  length,  one  in  force,  or  two  in 
length,  one  in  mass,  and  minus  two  in  time. 

Power. — According  to  the  definition  of  power  (art.  311),  a  unit 
of  power  is  proportional  directly  to  the  unit  of  work  and  inversely 
to  the  unit  of  time ;  hence  if  P  denotes  unit  of  power,  the  dimen- 
sional equation  is 

p  =  w/T  =LFT-^  =L2MT-^ 

and  a  unit  power  is  one  dimension  in  length  and  force  and  minus  one 
in  time,  or  two  in  length,  one  in  mass,  and  minus  three  in  time.* 

C  4.  Applications  of  the  Theory  of  Dimensions. — A  knowledge  of 
the  theory  of  dimensions  is  probably  of  most  value  to  the  beginner 
as  a  help  to  a  clear  understanding  of  the  different  mechanical  quan- 
tities and  the  relations  between  them.'  The  theory  is  useful  prac- 
tically in  other  ways,  two  of  which  we  mention." 

(i)  As  a  test, of  ike  accuracy  of  equations  between  mechanical  quan- 
tities.— Such  an  equation  if  rationally  and  correctly  deduced  must 
be  homogeneous,  i.e.,  the  terms  in  it  must  be  the  same  in  kind.     To 

*  Determination  of  the  dimensions  of  the  other  units  mentioned  in  the 
following  tables  is  left  to  the  student. 


360 


APPENDIX  C. 


ABSOLUTE  SYSTEMS  ("SCIENTIFIC"). 


Names  of  Quantities. 


Dimen- 
sional 
Formulas. 


Names  of  Units. 


C.G.S. 


P.P.S. 


Length , 

Mass , 

Time 

Velocity 

Acceleration 

Angular  Velocity 

Angular  Acceleration.  . . . 

Force 

Weight 

Moment  of  Mass 

Moment  of  Inertia  (Body) 

Moment  of  Force 

Work , 

Energy , 

Power , 

Impulse , 

Momentum 

Density 

Specific  Weight 

Moment  of  Area 

Moment  of  Inertia  (Area) 

Stress , 

Stress  Intensity 


L 

M 

T 

LT-' 

LT-' 
7-1 

LMT- 

LMT- 

LM 

UM 

UMT- 

L^MT- 

L^MT- 

L^MT- 

LMT- 

LMT- 

L-'M 

L-HiT- 

IJ 

L* 

LMT- 

L-'MT- 


centimeter  (cm) 

gram  (gr) 

second  (sec) 

cm /sec  ("kine") 

cm/sec^  ("spoud") 

rad/sec 

rad/sec^ 

dyne 

dyne 

gr-cm 

gr-cm 

cm-dyne 

cm-dyne  ("erg") 

cm-dyne  ("erg") 

erg /sec 

dyne-sec  ("bole") 

dyne-sec  ("bole") 

gr/cm^ 

dyne/cm^ 

cm^ 

cm* 

dyne 

dyne /cm'' 


foot  (ft) 

pound  (lb) 

second  (sec) 

ft /sec 

ft/sec2 

rad/sec 

rad/sec^ 

poundal  (pdl) 

pdl 

Ib-ft 

Ib-ft 

ft-pdl 

ft-pdl 

ft-pdl 

ft-pdl /sec 

pdl-sec 

pdl-sec 

lb/ft3 

pdl/ft3 

ft3 

ft* 

pdl 

pdl/ft2 


ascertain  whether  terms  are  the  same  in  kind  we  write  the  dimen- 
sional form  of  the  equation,  reduce  the  tenns  to  their  simplest 
forms  and  compare;  if  they  are  alike,  the  terms  are  the  same  in 
kind.     To  illustrate,  consider  eq.  (i),  art.  250, 

dt' 


^,,^r 


q^A  sin  cot,       .....      (i) 


in  which  y  and  A  denote  lengths,  t  and  x/p  time,  and  q  and  o)  angu- 
lar velocity.  Then  d'^y/dP  is  an  acceleration  and  dy/dt  a  velocity, 
and  the  dimensional  equation  is 

LT-'  +  {T-')(LT-')+(T-'yiL)  =  {T-'yL. 

An  abstract  number,  as  the  sine  of  an  angle,  is  independent  of  all 
■units  and  hence  does  not  affect  a  dimensional  equation.  Reducing 
the  terms  in  the  last  equation  we  get 

LT-2  -hLT-'  +LT-2  =LT-«; 

i.e.,  the  terms  are  alike  and  the  original  equation  is  homogeneous. 
Showing  that  the  equation  is  homogeneous  does  not  prove  that 


i  • 


DIMENSIONS   OF  UNITS 


361 


GRAVITATION  SYSTEMS  ("ENGINEERS'"). 


Names  of  Quantities. 


Dimen- 
sional 
Formulas. 


Names  of  Units. 


F.P.  (force)  S. 


M.K.  (force)  3. 


Length.  . 

Force.  .  . .  , 

Time 

Velocity. 

Acceleration 

Angular  Velocity.  . . . 
Angular  Acceleration. 

Mass 

Weight 

Moment  of  Mass.  ... 
Moment  of  Inertia. . . 
Moment  of  a  Force.  . 

Work , 

Energy , 

Power 

Impulse 

Momentum 

Density , 

Specific  Weight , 

Moment  of  Area.  .  .  .  , 
Moment  of  Inertia. . .  . 

Stress ^ , 

Stress  Intensity , 


L 

F 

T 
LT--" 
LT-^ 

L-'FT^ 
F 

LFT^ 

LF 

LF 

LF 
LFT-' 

FT 

FT 
L'FT^ 
L-W 

L' 

L* 

F 
L-'F 


foot  (ft) 

pound  (lb) 

second  (sec) 

ft /sec 

ft/sec2 

rad/sec 

rad/sec^ 

'geepound"  (gib) 

lb 

glb-ft 

glb-ft2 

ft-lb 

ft-lb 

ft-lb 

ft-lb /sec 

lb- sec 

lb-sec 

glb/ft' 

Ib/ft^ 

ft^ 

ft* 

lb 

lb/ft2 


meter  (m) 

kilogram  (kg) 

second  (sec) 

m/sec 

m/sec^ 

rad/sec 

rad/sec^ 

'geekilogram"  (gkg) 

kg 

gkg-m 

gkg-m^ 

m-kg 

m-kg 

m-kg 

m-kg/sec 

kg-sec 

kg-sec 

gkg/m* 

kg/m» 

m' 

m* 

kg 

kg/m» 


it  is  correct,  but  that  it  may  be  correct ;  showing  that  an  equation  is 
non-homogeneous  shows  it  to  be  incorrect.  Since  abstract  num- 
bers do  not  appear  in  the  dimensional  form  of  an  equation,  the  test 
for  homogeneity  does  not  discover  errors  in  numerical  coefficients 
and  terms,  nor  of  course  errors  in  signs. 

As  another  illustration  consider  eq.  (2),  art.  250, 


A  sin  (cot  +  e) (2) 


It  was  deduced  from  eq.  (i)  by  two  integrations.  If  we  find  that 
(2)  is  non-homogeneous,  we  may  be  sure  that  a  mistake  was  made  in 
the  deduction  from  (i)  to  (2).     The  dimensional  form  of  (2)  is 


(T-^)(T-^)  * 

i.e.,  the  equation  is  homogeneous  and  it  may  be  correct. 

Not  only  must  rational  mechanical  equations  be  homogeneous, 
but  every  factor  or  expression  in  it  which  is  the  sum  of  several  terms 
must  also  be  homogeneous.     For  example,  in  eq.  (2)  the  expression 


362  ^  APPENDIX  C. 

{(ot+e)  must  be  homogeneous,  and  since  e  denotes  an  angle,  it  is 
readily  seen  to  be  so. 

(2)  To  express  a  magnitude  in  different  units. — Obviously  the 
numerical  value  of  a  given  quantity  changes  inversely  as  the  mag- 
nitude of  the  unit  used ;  thus  a  certain  distance  may  be  expressed  as 

10  mi.,  17,600  yds.,  and  52,800  ft., 

and  plainly  the  numerics  are  respectively  as  i,  1760,  and  5280,  while 
the  corresponding  units  are  as  5280,  1760,  and  i. 

Let  q^  be  the  known  numerical  value  of  a  quantity  when  ex- 
pressed in  the  unit  Q^,  and  q^  the  numeric  (to  be  found)  of  the 
same  quantity  expressed  in  the  unit  Q^ ;   then 

qi  /^j  ==  Q2  /Qi  ^     or     q,=  q^Q,  /Q,. 

The  ratio  Q1/Q2  can  be  easily  computed  by  substituting  for  Q^  and 
Q2  their  equivalents  in  terms  of  fundamental  units ;  thus  if  a,  6,  and 
c  are  the  dimensions  of  Q^  (and  Q2), 

Q,^k,mM\n)     and     Q2  =  UKM\r,), 

wheie  Lj,  M^,  and  T^  are  the  particular  fundamentals  for  Qi.  -^2>  ^^zy 
and  7^2  those  for  Q2.  and  k^  and  k^  numerical  coefficients  (very  often 
unity).     Finally, 

As  an  example,  let  us  determine  how  many  watts  in  10  horse- 
power. Since  Q^  (horse-power)  =550  ft.-lb.-sec.~^  and  Qj  (watt)  =» 
10'  ergs  per  sec.  =  10^  cm.-dyne-sec.~S 

550  ft.      lb.    sec~^         550,         _.  .  .     5x/  X 

^' - '°x-f^  3Si:  d^  iiFT  = '°ft^  (30.48)  (4.45  X  io^)(r)  =  7640. 

(3)  To  ascertain  the  unit  of  the  result  of  a  numerical  calculation  — 
Substitute  for  the  quantities  the  names  of  the  units  in  which  they 
are  expressed,  and  then  repeat  the  calculation,  treating  the  names 
as  though  they  were  algebraic  quantities.  The  reduced  answer 
is  the  name  of  the  tmit  of  the  numerical  answer.  Thus  in  the 
formula  for  the  elongation  of  a  rod  due  to  a  pull  at  each  end,  Pl/AE, 
wherein  P  denotes  pull,  /  length  of  the  rod,  A  area  of  cross-section, 
and  E  Young's  modulus  for  the  material,  suppose  that  P=  10,000 
lbs.,  /  =  5o  in.,  ^  =  0.5  in.^,  £^  =  30,000,000  Ibs./in.^;  the  calculations 
for  elongation  *id  name  of  unit  are 

10,000X50      _-„       „_^     lbs.  X  in.      _  lbs.  X  in.  X  in.'  _ . 

0.5X30,000,000       •^^'  in.^Xlbs./in.^        in.^Xlbs. 


APPENDIX  D. 

SECOND  MOMENTS  OF  AREAS  (MOMENT  OF  INERTIA,  ETC.).* 

In  the  subject  of  strength  of  materials  especially,  the  student  of 
engineering  meets  with  quantities  expressed  by  integrals  of  the  kind 

and  form  /  dA-x^  and  /  dA  -xy,  A  denoting  area  and  x  and  y  dis- 
tance. Such  quantities  have  been  called  "moments  of  area  of  the 
second  order,"  or  briefly  "  second  moments  of  area,"  the  terms  being 
in  line  with  "  first  moments  of  area,"  which  term  is  applied  to  quan- 
tities expressed  by  integrals  like  J  dA'X  (see  art.  83).  We  distin- 
guish between  second  moments  of  area  employing  special  names 
for  the  kinds. 

§  I.     Moment  of  Inertia. 

D  I.  Moment  of  Inertia  Defined.  —  The  moment  of  inertia  of  a 
plane  area  with  respect  to  any  axis  is  the  sum  of  the  products 
obtained  by  multiplying  each  elementary  part  of  the  area  by  the 
square  of  its  distance  from  the  axis. 

The  axis  of  reference  will  often  be  called  "inertia-axis"  to  dis- 
tinguish it  from  other  axes,  coordinate,  geometrical,  etc.  We  con- 
sider only  moments  of  inertia  with  respect  to  axes  in  or  normal  to 
the  plane  area;  the  latter  are  called  polar  moments  of  inertia,  and 
the  corresponding  axes  polar  axes. 

Expression  for  Moment  of  Inertia. — Let  dA^,  dA^,  dA^,  etc.,  de- 
note elementary  parts  of  an  area  and  p^,  p^,  p^,  etc.,  respectively 
their  distances  from  some  axis;  then  according  to  the  definition, 
the  moment  of  inertia  of  the  area  with  respect  to  that  axis  is 

{dA,)p^  +  {dA,)p-^  +  .., 

Or,  if  7 1  denotes  the  moment  of  inertia,  dA  any  elementary  por- 

*  Writers  on  Strength  of  Materials  usually  refer  to  works  on 
Mechanics  for  a  treatment  of  these  second  moments,  and  for  that  reason 
this  appendix  is  herein  included. 

t  A  subscript  affixed  to  the  symbol  refers  to  the  inertia-axis;  thiis7a 
stands  for  moment  of  inertia  with  respect  to  the  x  axis. 

363 


364  APPENDIX  D, 

tion  of  the  area  all  points  of  which  are  equally  distant  from  the 
axis,  and  p  that  distance, 


fdA'p\ 


[The  term  moment  of  inertia  of  an  area  is  unfortunate  because 
beginners  are  prone  to  seek  reasons  for  its  appropriateness  which 
do  not  exist.  -  They  should  recognize  at  the  outset  that  an  area 
has  no  inertia  and  hence,  in  the  ordinary  sense  of  the  words,  no 
moment  of  inertia.  The  reason  why  this  second  moment  of  an 
area  was  so  called  lies  in  the  fact  that  the  moment  is  closely  "analo- 
gous to  another  quantity  (a  second  moment  of  mass,  see  art.  254) 
which  had  previously  been  called  moment  of  inertia  of  a  body.] 

D  2.  Units  of  Moment  of  Inertia. — Each  term  in  the  preceding  series 
is  the  product  of  four  lengths ;  hence  a  moment  of  inertia  of  an  area 
is  four  "  dimensions  "  in  length.  The  numerical  value  of  a  moment 
of  inertia  of  an  area  is  usually  computed  with  the  inch  as  unit 
length,  and  the  corresponding  unit  moment  of  inertia  is  called  a 
"biquadratic  inch,"  abbreviated  thus:  in.* 

D  3.  Radius  of  Gyration. — Since  any  moment  of  inertia  of  an 
area  is  four  "dimensions"  in  length,  it  can  be  expressed  as  the 
product  of  an  area  and  a  length  squared.  It  is  sometimes  con- 
venient to  so  express  it. 

Definition. — The  radius  of  gyration  of  an  area  with  respect  to 
an  axis  is  such  a  length  whose  square  multiplied  by  the  area  equals 
the  moment  of  inertia  with  respect  to  that  axis.  That  is,  if  k  and 
/  denote  the  radius  of  gyration  and  moment  of  inertia  of  an  area  A 
with  respect  to  the  same  axis, 

k2A=I,     or     k  =  V'l7A. 

The  square  of  the  radius  of  gyration  of  an  area  with  respect  to 
an  axis  i*^  the  mean  of  the  squares  of  the  distances  of  all  the  equal 
elementary  parts  of  the  area  from  that  axis.  For  let  p^,  p^,  etc., 
be  the  distances  from  the  elements  {dA)  to  the  axis,  and  let  n  de- 
note their  number  (infinite) ;  then  the  mean  of  the  squares  is 

W-^Pz'^Pi^  ...)/n  =  {p;'dA+p,'dA+  ,,.)/ndA=I/A. 

But  I /A  is  the  square  of  the  radius  of  gyration,  hence,  etc. 

[Like  moment  of  inertia,  the  term  radius  of  gyration  when  ap- 
plied to  areas  is  strictly  inappropriate.  Its  use  in  'this  connection 
is  justified  by  analogy,  the  quantity  k  being  closely  analogous  to 
another  quantity  which  had  been  previously  called  radius  of  gyra- 
tion (art.  255).] 


SECOND  MOMENTS  OF  AREAS. 


365 


EXAMPLES. 

1.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
rectangle  with  respect  to  a  central  axis  parallel  to  the  base  are  respect- 
ively 

iV^a'     and     aVi/12, 
b  and  a  denoting  the  base  and  altitude  respectively. 
Solution:  We  will   take  a  horizontal  strip   as 
elementary  area  (see  fig.  D  i) ;  then  dA  =bdy,  and 

/x=  /  bdy-y'^  =  ^^ba^. 

t/— a/2 

Also,  A=bay  therefore  ^*  = -^60 V^ct,  or fe  =a\/i/i 2. 

2.  Show  that  the  moment  of  inertia  and  radius 
of  gyration  of  a  triangle  with  respect  to  a  central 
axis  parallel  to  the  base  are  respectively 

i-^bd^     and     aVi/iS, 
b  and  a  denoting  the  base  and  the  altitude  respectively. 

Solution:  We  will  take  a  horizontal  strip  as  elementary  area 
(fig.  D  2),  calling  the  length  of  any  strip  u,  then 
dA  =udy\  also, 


br\ — ^- — 


Fig.  D  I. 


t^_la—y. 


Fig.  D2. 


b 
Hence 


therefore  dA^{2/^—y/a)bdy. 


,2  a/ 3 


f       (2/s-y/a)bdyy*^Ma*. 

J -a/ 3 


Also,  since  A  =^ba,  k^  '^Ma'/hba   or  k  =a^i/iS. 

3.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
circle  with  respect  to  a  diameter  are  respectively 

^Tir*     and     ^r, 
r  denoting  radius  of  the  circle. 

Solution:    We  will  take  a  horizontal  strip  as  elementary  area 
(see  fig.  D  3  a) ;  then  if  u  denotes  the  length  of  any  strip, 

2  V? 


dA  =udy 


■y^dy. 


Hence 
and  since  A 


Vr^  —y^dy'y^  =  l7ir*; 


nr' 


k^  =  \7zr*/r.r'^,     or    kx^h'f- 
4.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
circle  with  respect  to  a  central  polar  axis  are  respectively 

i;rr*    and     rVi/a. 


366 


APPENDIX  D. 


Solution:   We  choose  as  elementary  area  one  as  represented  in 
fig.  D  3  (6) ;  then  dA  =  pdMpy  and 


/«=  /""  ripdd'dp)p^^i7:r\  etc. 

Jo  V  o 


In  the  preceding  examples,  the  inertia-axes  are  central,  but  mo- 
ments of  inertia  of  such  areas  can  be  readily  computed  by  integra- 


FiG.  D  3. 

tion  for  any  axes  which  are  simply  situated  with  reference  to  a 
line  of  the  figtire.  Such  cases  are  rectangle  and  triangle  with  inertia- 
axis  parallel  to  a  side,  circle  with  inertia-axis  parallel  to  a  diam- 
eter, etc.  In  the  next  article  it  is  shown  how  to  determine  moments 
of  inertia  without  integration  in  these  and  similar  cases,  but,  to  test 
his  understanding  of  the  integration  method,  the  student  should 
determine  the  moment  of  inertia  of 

(o)  a  rectangle  with  respect  to  a  side ; 

(6)   "  triangle        "  "         "  "   "    ; 

(c)    "  circle  "  "  "  "  tangent. 

D  4.  Relations  between  Moments  of  Inertia  and  between  Radii  of  Gyra- 
tion with  Respect  to  (a)  Two  Parallel  Axes,  (b)  Three  Rectangular  Axes. — 
Proposition  I. — The  moment  of  inertia  (/)  of  an  area  with  respect 
to  any  axis  equals  its  moment  of  inertia  (/)  with  respect  to  a  parallel 
centroidal  axis  plus  the  product  of  its  area  {A)  and  the  square  of 
the  distance  {d)  between  the  axes,  or,  symbolically, 

I=I+Ad2 ".     .     .     .     (i) 

Proof:  (a)  The  inertia-axis  is  in  the  plane  area.  Let  the  area 
be  that  represented  in  fig.  D  4  (a),  U  being  the  inertia  axis  and  C 
the  centroid.     Then 

I  =  JdA'v'  =  fdA(y  +dy  =  f  dA  y'  +2dfdA'y  +d^fdA, 
But    fdA'y^-=I,    jdA'y=Ay=-o,      and     fdA^A\    hence,  etc. 


SECOND  MOMENTS  OF  AREAS. 


367 


(6)  The  inertia-axis  is  normal  to  the  area.  Let  the  area  be  that 
represented  in  fig.  D  4  (6),  O  the  point  where  the  inertia-axis  pierces 
the  plane  of  the  area,  and  C  the  centroid.     Then 

I  =  fdAiy'  +^^')  =  f  dA(y^  +x')  +2d  fdA'X  +d'fdA. 
Now 

fdA(y^+x^)=T,     fdA'X=Ax  =  o,    and    jdA=-A\    hence,  etc. 

Corollary:  Dividing  both  sides  of  eq.  (i)  by  A,  we  have 

/M=7M+d^     or    k2  =  k'+d2 (2) 

h  denoting  the  radius  of  gyration  of  the  area  with  respect  to  any  axis, 
and  k  that  with  respect  to  a  parallel  centroidal  axis. 

Y 


(a)  (6) 

Fig.  D4. 

Equations  (i)  and  (2)  show  that  the  moment  of  inertia  and 
radius  of  gyration  of  an  area  with  respect  to  a  centroidal  axis  are 
respectively  less  than  those  for  any  other  parallel  axis.  Also,  that 
with  respect  to  any  axis  the  radius  of  gyration  {k)  is  always  greater 
than  the  ordinate  {d)  from  that  axis  to  the  centroid,  but  if  k  is  small 
compared  to  d^ 

k=d     and     I  =  Ad^,  approximately. 

Proposition  II. — A  polar  moment  of  inertia  of  an  area  (J^  equals 
the  sum  of  its  moments  of  inertia  {Ix  and  ly)  with  respect  to 
any  rectangular  axes  in  the  area  which  intersect  the  polar  axis,  or 

Iz=Ix+Iy (3) 

Proof:  In  accordance  with  the  notation  above,  the  inertia-axes 
in  the  plane  of  the  area  must  be  called  x  and  y  coordinate  axes  and 
the  polar  one  the  z  axis.  Then  the  distance  of  any  point  of  the 
area  from  the  z  axis  is  \/x^+y^,  and  therefore 

Iz  =  fdA(x^  +y')  =  fdA'x'  +  fdA'y\ 

But  jdA 'X^=Iv    and    I dA'y^  =  Ix\  hence,  etc. 


368  ,  APPENDIX  D. 

Corollary. — Dividing  both  sides  of  eq.  (3)  hy  A,  we  get 

h/A=Iz/A+Iy/A,     or     kj=ki+kj,       ...     (4) 

ke  denoting  the  radius  of  gyration  with  respect  to  the  polar  axis, 
kx  and  ky  those  with  respect  to  any  rectangular  axes  in  the  plane  of 
the  area  cutting  the  polar  one. 

Equations  (3)  and  (4)  show  that  the  sums  Ix+Iy  and  kl+kl 
are  the  same  for  all  directions  of  the  axis  x  (and  y).  Hence  if  we 
imagine  the  x  and  y  axes  (90°  apart)  to  turn  about  the  polar  axis, 
when  Ix  and  kx  reach  a  maximum  or  minimum  value,  ly  and  ky  are 
a  minimum  or  maximum. 

EXAMPLES. 

1.  Show  without  integration  that  the  moment  of  inertia  and 
radius  of  gyration  of  a  rectangle  with  respect  to  its  base  are  respect- 
ively 

^ba^     and     aVTJ^, 

6  and  a  denoting  the  base  and  altitude  (see  ex.  i,  art.  D  3). 

2.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
triangle  with  respect  to  its  base  are  respectively 

yV6a^     and     aVi/6, 

h  and  a  denoting  base  and  altitude  respectively  (see  ex.  2,  art.  D  3). 

3.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
square  with  respect  to  a  central  polar  axis  are  respectively 

|6*     and     bVTj6, 

b  denoting  the  length  of  its  sides. 

4.  From  the  result  of  ex.  3,  art.  D  3,  deduce  the  expression  for 
the  moment  of  inertia  of  a  circle  with  respect  to  a  central  polar  axis. 

5.  Show  that  the  moments  of  inertia  of  a  square  with  respect  to 
a  diagonal  and  a  central  axis  parallel  to  any  side  are  equal. 

6.  Compute  the  moments  of  inertia  and  radii  of  gyration  of  a 
rectangle  i  X  12  in.  with  respect  to  axes  7  inches  from  the  centre  and 
parallel  to  the  sides.  Compare  the  radii  of  gyration  with  the  dis- 
tance from  the  axes  to  the  centre.       Ans.  Greater /==  732  in.* 

D  5.  Composite  Areas. — We  refer  now  to  areas  which  can  be 
divided  into  simple  component  parts;  e.g.,  a  trapezoid  divisible  into 
two  triangles,  a  circular  annulus,  consisting  of  a  circle  minus  a 
smaller  one,  etc.  The  moment  of  inertia  of  such  an  area  with 
respect  to  any  axis  can  be  computed  by  adding  algebraically  the 


SECOND  MOMENTS  OF  AREAS. 


2>'^:i 


moments  of  inertia  of  its  parts  with  respect  to  the  same  axis,  the 
moments  of  inertia  of  the  "  negative  component  parts  "  being  given 
the  minus  sign. 

EXAMPLES. 

I.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
circular  annulus  with  respect  to  a 
diameter  (fig.   D  5  a)    are  respect- 
ively 
i7r(r,*  -r/)     and      ^Vr^TV. 
Solution :  The  moment  of  iner- 


•>— 6r-> 

I 

-i- 

I 
I — i 


tia  of   the   larger   circle   is   ^nr^'' 

(see  ex.  3,  art.  D  3)  and  that  -of         ""       ^  *      bi — > 

the  smaller  is  i^rr^*.     Hence  the  ^^^  (&) 

moment  of  inertia  of  the  annulus  Fig.  D  5. 

is  i7r(rj*— r^*),  etc. 

2.  Show  that  the  radius  of  gyration  of  a  circular  annulus  with 
respect  to  a  central  polar  axis  is  "^ {r^  -\-r^^) /i. 

3.  Show  that  the  moment  of  inertia  and  radius  of  gyration  of  a 
hollow  rectangle  with  respect  to  a  central  axis  parallel  to  the  base  (fig. 
D  5  6)  are  respectively 

Tif(^2«2'  -^i^i')     and     {{b^a^^  -b^a^^) /i2{b^a.,  -b^a^)yt. 

4.  Compute  the  moments  of  inertia  of  the  "angle"  section  of  fig. 
D  6  with  respect  to  the  x  and  y  axes  respectively. 

Solution:  Consider  that  the  section  consists  of  the  rectangle 
A  BCD  minus  the  rectangle  A'B'C'D.  The  moments  of  inertia  of 
these  with  respect  to  the  line  CD  are  (see  ex.  i,  art.  D  4) 

i  3-5  X7^  =  40o.i6     and     ^  2.5  X6'  =  180  in.* 

Hence  the  moment  of  inertia  of  the  section  with  respect  to  CD  is 
400.16—180  =  220.16,  The  area  of  the  section  being  9.5  in.*,  the 
moment  of  inertia  sought  is  (see  eq.  (i),  art.  D  4), 

220.16 -9.5(7 -2.71)2  =  45.32  in.* 

5.  Show  that  for  the  Z  section  of  fig.  D  6 

I^  =  ^^Xha' -{b-t){a-2tyi 

6.  Deduce  an  expression  for  the  moment  of  inertia  of  the  T 
section  of  fig.  D  6  with  respect  to  a  central  axis  parallel  to  the 
base. 


370 


APPENDIX  D. 


7.  Compute  the  moment  of  inertia  of  the  area  represented  in 
fig.  D  7  (a  section  of  a  "  built-up "  steel  beam)  consisting  of  a 
"web    plate,"   two   "side  plates,"  an(i  four  Z  bars)  with   respect 


i' 


c   I 


I  <^-K-i- 


1° 


^ajrt? 


^i. 


(a) 


(6) 
Fig.  D6. 


Y 

1 

! 

i 

r 

w 

1    1: 

':'   1 

—b- 
(c) 


6%" 


to  a  horizontal  central  axis,  the  moment  of  inertia  of  one  Z  section 
with  respect  to  its  horizontal  central  axis  being  given  as  50.22  in.* 

(Such  numbers  and  similar  data  are  ob- 

•<- ■i8^- ->  tainable  from  "  handbooks  "  published 

'  ■'  ■  ■  ^  by  steel-manufacturing  companies.) 

Solution:  We  compute  the  moment 
of  inertia  of  the  parts  separately.  Side- 
plate  section:  moment  of  inertia  with 
respect  to  its  horizontal  central  axis  = 
tV  18X1^  =  1.5  in.^;  area  =  i8  in.^; 
distance  from  its  centroid  to  the  inertia- 
axis  specified  =  7!-  in.  Hence  for  two 
side  plates 

/x  =  2(1.5  + 18  X  7F)  =  1830.52  in.* 
l*iG.  D  7.  Z    section:     moment    of    inertia    with 

respect  to  its  horizontal  central  axis  =  50.22  in.*;  area  =  10.17  in.'; 
distance  between  its  centroid  and  the  inertia-axis  —  3yV  in. ;  hence 
for  the  four  Z  bars 

/ai  =  4(5o.22  +10.17  x^-^if '0  =  717-16  in.* 
Web-plate  section:    moment  of  inertia  with  respect  to  the  x  axis  is 

tV  10X1^  =  0.83  in.* 
For  the  entire  section: 

7x  =  i83o. 52  +  717. 16  +  0. 83-2548. 51  in.' 


SECON<D  MOMENTS.  OF  AREAS.  371 

8.  The  moment  of  inertia  of  one  Z  section  (fig.  D  7)  with 
respect  to  its  vertical  central  axis  being  19.18  in.*,  compute 
the  moment  of  inertia  of  the  composite  area  with  respect  to  the 
y  axis. 

9.  The  radii  of  gyration  of  the  "angle"  of  fig.  D  6  with  re- 
spect to  its  horizontal  and  vertical  central  axes  are  2.19  and 
0.89  in.  respectively.  Compute  the  radii  of  gyration  of  a  pair 
of  such  sections  with  respect  to  their  horizontal  and  vertical  central 
axes,  they  being  placed  so  that  AB  is  an  axis  of  symmetry  for  the 
pair, 

§  II.     Product  of  Inertia. 

D  6.  Product  of  Inertia  Defined. — The  product  of  Inertia  of  a 
plane  area  with  respect  to  a  pair  of  coordinate  axes  in  the  plane  is 
the  sum  of  all  the  products  obtained  by  multiplying  each  ele- 
mentary area  by  its  coordinates. 

Expression  for  Product  of  Inertia. — Let  dA^,  dA^,  dA^,  etc.,  de- 
note elementary  parts  of  an  area  and  {xi,  y^),  (x^,  ^'2).  etc.,  their 
coordinates  respectively ;  then,  according  to  the  definition,  the  prod- 
uct of  inertia  of  the  area  with  respect  to  the  coordinate  axes  is 

{dA,)x,y^  +  (dA,)x,y,  +etc., 

or,  if  /  *  denotes  the  product  of  inertia,  dA  any  element  of  the  area 
(dA  must  be  of  the  second  order,  as  dx  dy) ,  and  x  and  y  its  coordi- 
nates, then 


J  =  /dA.xy, 


the  limits  of  integration  ^being  assigned  so  that  the  products  d A xy 
for  all  elements  are  included  in  the  integration. 

D  7.  Units  of  Product  of  Inertia. — It  is  plain  from  the  definition 
and  expression  of  the  preceding  article  that  a  unit  product  of  inertia 
is  four  "  dimensions"  in  length.  Like  moments  of  inertia  we  will 
express  products  of  inertia  in  biquadratic  inches. 

EXAMPLES. 

I.  Deduce  an  expression  for  the  product  of  inertia  of  the  rect- 
angle (fig.  D  8)  with  respect  to  the  coordinate  axes. 

Solution:  J  =  JdA'Xy=  f  J^'idx  dy)xy  =  Wa'  =  iA\ 

*  When  it  is  necessary  to  specify  the  axes  with  respect  to  which  the 
product  of  inertia  is  taken,  they  are  indicated  by  subscripts  thus:  Jxy. 


72 


APPENDIX  D. 


y 

4 
1 

— 

-.^...^dA 

a 

i 

1 

i 

1 

1 

1 

-      1 

X 

—6 > 

Fig.  D8. 


2.  Deduce  expressions  for  the  products  of  inertia  of  the  rect- 
angle with  respect  to  coordinate  axes  through 
the  other  corners^and  centre. 

D  8.  Zero  Products  of    Inertia.  —  UnHke     a 
moment  of  inertia,  a  product  of  inertia  may- 
be negative  or  zero.     The  pair  of  -axes  pass- 
ing through  a  given  point  with  respect  to 
which  the  product  of  inertia  of  an  area  equals 
zero  is  often  of  practical  importance ;  in  art. 
D  14  a  method  is  given  for  finding  such  axes. 
The  following  proposition  enables  one  to  find 
those  axes  more  readily  in  certain  cases. 
Proposition. — If  an  area  has  an  axis  of  symmetry,  its  product 
of  inertia  with  respect  to  that  axis  and  any  other  perpendicular  to 
it  is  zero. 

Proof:  Let  the  elementary  parts  of  the  area  be  grouped  into 
pairs,  the  elements  of  each  pair  being  symmetrically  situated  with 
reference  to  the  axis  of  symmetry.  Obviously  the  product  of  iner- 
tia of  each  pair  is  zero,  and  hence  the  product  of  inertia  of  all  the 
pairs  (the  entire  area)  is  also  zero. 

EXAMPLES. 

I.  With  respect  to  which  axes  are  the  products  of  inertia  of  the 
following  zero:  (i)  rectangle;  (2)  isosceles  triangle;  (3)  circle: 
(4)  "  channel,"  "  tee,"  and  "  angle  of  equal  legs  "  (fig.  49)? 

D  9.  Relation  between  Products  of  Inertia  for  Parallel  Pairs  of  Axes, 
Proposition. — The  product  of  inertia  (J)  of  an  area  with  respect  to 
any  pair  of  coordinate  axes  equals  its  product  of  inertia  (/)  with 
respect  to  a  parallel  pair  of  central  axes  plus  the 
area  {A )  times  the  product  of  the  coordinates  of 
the  centroid  {x,  y)  referred  to  the  first  set  of 
axes,  or 

J=J+Axy.        .     .     .     .     (i) 

Proof:    Let  OX  and  OY   (fig.  D  9)  be  the 
axes  and  C  the  centroid.     Then 


/  =  /  dA  •  xy,     and     /  =  /  dA  •  mv. 
Since     x  =  u+x  and  y=v  -\-y, 


Y  ^ 

r?) 

,  u    1        J  \i 

0 

'W 

v-^ 

Fig.  Dp. 


J  =  jdA  {u+x){v+y)=  JdA  •  uv  +  yjiidA  +  xfvdA  +  xyjdA. 
But   j  udA  =uA  =0,  J  vdA  =vA=o,  and  J  dA'^A;  hence,  etc. 


SECOND  MOMENTS  OF  AREAS. 


373 


n 

cj 

1   r 

hojA       I 

D  10.  Composite  Areas. — The  products  of  inertia  of  such  an  area 
with  respect  to  any  pair  of  coordinate  axes  equals  the  algebraic  sum 
of  the  products  of  inertia  of  its  component  parts  with  respect  to 
the  same  axes. 

EXAMPLES. 

1 .  Compute  the  product  of  inertia  of  the  angle  section  of  fig.  D  6 
with  respect  to  axes  OX  and  0  Y. 

Solution:    Divide  the  figure  into  two  rectangles  as  shown  in  fig. 
D  lo ;  the  areas  of  the  parts  are  3.5  and  6  in.',  and 
their  centroids  are  at  C^  and  C^  respectively,  C 
being  the  centroid  of  the  entire  figure. 

For  the  first  rectangle  the  product  of  inertia 
with    respect    to    horizontal    and    vertical    axes 
through  Ci  equals  zero  (see  prop.,  art.  D  8),  and 
hence,  according  to  the  preceding  article, 
/xy  =  o+3.5(o.79)(-2.2i)=  -6.11  in.* 

For  the  second  rectangle  the  product  of  in- 
ertia with  respect  to  horizontal  and  vertical  axes 
through  C2  equals  zero,  and  hence 

/xy  =  o+6(-o.46)(i.29)=  -3.56  in.* 

For  the  entire  figure,  therefore, 

/xy  =  (-6.ii)  +(-3.56)=  -9.67  in.* 

2.  Show  that  the  product  of  inertia  of  the  Z  section  of  fig.  D  6 
with  respect  to  the  :*:  and  y  axes  is  —  11.54  in.*,  a,  b,  and  t  being  6, 
3^,  and  I  in.  respectively. 

§  III.     Relation  between  Moments  op  Inertia  with  Respect 
TO  Axes  Inclined  to  Each  Other. 

D  II.  General  Equations. — Let  OX  and  OY  (fig.  D  11)  be  any 
two  rectangular  axes  and  OU  and  OV  another  pair,  XOU  being  any 

angle  a.     From  the  figure  it  is  plain  that 

u=y  sin  a  +x  cos  a, 
and  v=y  cos  a  —x  sin  a. 

If  these  values  for  u  and  v  be  substituted 


Fig.  D  10. 


^v 

[Y 

\r 

aj--- 

'~j' 

^<^AV 

\ 

..^<r 

Uu 

\ 

.^-<^ 

^W    X 

\^ 0 

^ 

Fig.  D  II. 

it  will  be  found  that 


lu-'JdA'v'    and     J^^^fdA 


374  ^  APPENDIX  D. 

/m=/x  cos^a +/y  sin^a  — /xy  sin  2  a (i) 

and  Juv  =  h{Ix—Iy)  svcv  2a +Jxy  cos  2a (2) 

With  these  equations  it  is  possible  to  find  the  moment  of  inertia 
with  respect  to  any  axis  through  any  point  and  the  product  of  iner- 
tia with  respect  to  any  pair  of  axes  through  that  point,  if  the  mo- 
ments and  the  product  of  inertia  of  the  area  with  respect  to  two 
rectangular  axes  through  the  point  are  known. 

EX/^MPLES. 

1.  Take  u  and  v  axes  through  0  in  fig.  D  6  (a)  inclined  at  45° 
and  135°  to  the  x  axis,  and  compute  luSind  Juv,  it  being  given  that 
Ix,  ly,  and  J zy  equal  45.37,  7.53,  and   —9.67  in.*  respectively. 

Solution:   Substituting  in  eqs.  (i)  and  (2),  we  find  that 

-    /u  =  45-37  cos' 45° +7-53  sin' 45° +9-67  sin  90°  =  36.12  in.* 
and      /«v  =  i(45-37  —7-53)  sin  90°  —9.67  cos  90°  =  18.92  in.* 

2.  Consider  a  rectangle  whose  base  and  altitude  equal  4  and  6 
in.  respectively,  and  compute  its  moment  of  inertia  with  respect  to 
a  line  perpendicular  to  either  diagonal  at  its  end;  also  its  product 
of  inertia  with  respect  to  that  line  and  the  diagonal. 

D  12.  Geometrical  Constructions. — Equations  (i)  and  (2)  of  the  pre- 
ceding article  can  be  solved  graphically.  There  are  two  graphical 
methods;  in  one  a  certain  ellipse,  the  "  inertia-ellipse,"  is  employed, 
and  in  the  other  a  certain  circle,  which  will  herein  be  called  "  inertia- 
circle."  The  former  possesses  more  elegance  and  is  probably  more 
powerful,  but  the  latter  is  as  much  simpler  to  apply  as  is  the  draw- 
ing of  a  circle  compared  to  that  of  an  ellipse.  Only  a  brief  discus- 
sion of  the  inertia-circle  method  can  be  given  herein.* 

D  13.  Inertia- Circle. — Let  it  be  required  to  ascertain  the  moments 
and  product  of  inertia  of  the  area  of  fig.  D  1 2  with  reference  to  any 
coordinate  axes  u  and  v  through  the  point  O.  Let  Ox  and  Oy  be 
two  axes  with  respect  to  which  the  moments  and  product  of  inertia 
of  the  area  are  known.     Then  lay  off 

OX  =  Ix,     OY  =  fy,     and     YA=Jxy, 

OX  and  OF  along  the  positive  x  axis  and  YA  in  the  positive  or  nega- 
tive y  direction  according  as  Jxy  is  positive  or  negative.     The  circle 

*  The  invention  of  the  inertia-circle  is  due  to  Prof.  Culmann.  For^a 
full  treatment  see  his  "  Graphische  Statik"  or  that  of  Muller-Breslau,  or 
a  paper  by  Prof.  L.  J.  Johnson  in  the  Jour.  Assoc.  Eng.  Soc,  No.  5,  Vol. 

xxvin. 


SECOND  MOMENTS   OF  AREAS. 


375 


whose  centre  is  midway  between  X  and  Y  (the  point  C)  and  passing 
through  A  is  the  inertia-circle  for  the  area  corresponding  to  the 
axes  X  and  y. 


(G)  (d) 

Fig.  D  12. 
To  determine  /«  and  Juv- — From  A  draw  a  secant  parallel  to  the 
u  axis,  thus  determining  a  point  B,  and  from  5  drop  a  perpendicular 
to  the  X  axis,  thus  determining  a  point  U.     Then 

OU  =  Iu     and     UB=Juv 
to  the  scale  used  in  laying  ofT  OX,  OY,  and  F^.*  /«t,  is  positive  or 

*  Proof:  In  part,  fig.  D  1 3  is  a  reproduction  of  a  portion  of  fig.  D  1 2  (a) ; 
XA'  is  made  equal  to  YA, hence  A'Bb 
is  perpendicular  to  AB  and  to  the  u 
axis.  The  construction  of  the  other 
dotted  lines  is  obvious,  they  being 
either  parallel  or  perpendicular  to  the 
uor  X  axis. 

Equation    (i),   art.   D   11    can    be 
written  thus: 

lu  =  {Ix  cos  oc—Jxy  sin  a)  cos  a 

+  (Iy  sin  a  —Jxy  cos  a)  sin  a. 


Fig.  D  13. 

and  this  expression  for  /«  can  be  readily  evaluated  from  the  figure, 
is  plain  that  since  OX  =Ix  and  XA'  =Jxy, 

Ix  cos  a  =0a,     Jxy  sin  a  =-ab,     Ix  cos  a- Jxy  sin  a  ^Ob, 
and  (Ix  cos  a— Jxy  sin  a)  cos  a=0c; 


It 


376 


APPENDIX  D. 


negative  according  as  UB  is  along  the  positive  or  negative  y  direc- 
tion. If  it  should  happen  that  Jxy  equals  zero,  the  construction 
would  simplify ;  for  in  that  case  A  coincides  with  Y,  and  hence  XY 
is  a  diameter  of  the  inertia-circle, 

EXAMPLES. 

I.  Solve  ex.  i,  art.  D  ii  by  means  of  the  inertia-circle. 
Solution:    Lay  off  on  the  x  axis  (fig.  D  14)  OX,  OY,  and  YA  to 


Fig.  D  14. 


Fig.  D  15. 


represent  Ix,  ly,  and  Jxy  (scale  i  in.  =30  in.*).  Then  draw  a  circle 
with  centre  at  C  and  radius  equal  to  CA ;  this  is  the  inertia-circle 
corresponding  to  the  axis  Ox  and  Oy.  Next  draw  a  line  through  A 
parallel  to  the  u  axis,  mark  its  intersection  with  the  circle  B,  and 
drop  the  perpendicular  BU]  then  OU  (1.2  in.)  and  UB  (0.62  in.) 
represent  the  moment  of  inertia  and  product  of  inertia  sought. 
2.  Solve  ex.  2,  art.  D  11  by  means  of  the  inertia-circle. 

also  that,  since  Oy=Iy  and  YA  =Jxy, 

ly  sin  a  =  Yd,      Jxy  cos  a  =  Ye,     ly  sin  a—  Jxy  cos  a  =de  =Bh^ 
and  {ly  sin  a  —Jxy  cos  a)  sin  a=Bf  =cU. 

Hence  Iu=Oc  +  cU  =0U.  q.e.d. 

Equation  (2) ,  art.  D  1 1  can  be  written  thus : 

Juv  =  {Ix  cos  a  —Jxy  sin  a)  sin  a  —  {ly  sin  a—  Jxy  cos  a)  cos  a, 
and  this  expression  for  Juv  can  be  readily  evaluated  from  the  figure.     As 
already  explained, 

Iz  cos  a  —Jxy  sin  a—Ob, 
hence  {Ix  cos  ol—  Jxy  sin  a)  sin  a  =hc\ 

and  since,  as  explained,  ly  sin  a— Jxy  cos  a=Bh, 

{Ix  sin  a  —Jxy  cos  a)  cos  a  =bf. 
Hence  Jxy=bc  —  bf  =fc=BU.  q.e.d. 


SECOND  MOMENTS  OF  AREAS.  377 

D  14.  Principal  Axes  and  Principal  Moments  of  Inertia. — As  the  angle  a 
(fig.  D  12)  increases  from  o  to  360°,  the  point  U  moves  along  the  x 
axis,  its  extreme  positions  being  P^  and  P^,  and  for  these  two  posi- 
tions UB  equals  zero.     Hence 

(a)  the  maximum  and  minimum  values  of  /«  are  given  by  OPx 
and  OP2; 

(6)  the  corresponding  inertia-axes  are  parallel  to  AP^  and  AP.^ 
(therefore  rectangular) ; 

{c)   the  product  of  inertia  with  respect  to  those  axes  equals  zero. 

If  Ix  equals  ly  and  Jxy  =  o,  the  inertia-circle  vanishes,  and  hence 
lu  is  constant  and  Juv  equals  zero  for  all  values  of  a. 

Definitions. — The  two  axes  through  a  point  with  respect  to 
which  the  moments  of  inertia  of  an  area  are  greater  and  less  than 
for  any  other  axes  through  that  point  are  called  the  principal  axes  of 
the  figure  at  that  point,  and  the  corresponding  moments  of  inertia 
and  radii  of  gyration  are  called  the  principal  moments  of  inertia  and 
radii  of  gyration  of  the  area  at  that  point.  We  will  denote  these 
maximum  and  minimum  moments  of  inertia  and  radii  of  gyration 
by  /j,  Jj,  k^,  and  ^2  respectively,  and  will  mark  the  corresponding 
inertia-axes  (i)  and  (2)   (see  fig.  D  12). 

According  to  the  above  the  product  of  inertia  of  an  area  with 
respect  to  the  principal  axes  at  a  point  equals  zero.  This  proposi- 
tion leads  at  once  to  an  algebraic  method  for  finding  principal  axes 
and  moments  of  inertia.  Thus  let  oc'  denote  the  value  of  a  which 
makes  Juv  zero,  then  (see  eq.  (2),  art.  D  11) 

i(/z  —ly)  sin  2a'  +Jxy  cos  2a'  =  o, 
or  tan  2a' =  2 Jxy/ily—lx) (i) 

By  means  of  this  equation  we  may  locate  the  principal  axes  at  a 
point  (i.e.,  determine  a'),  and  then  determine  the  principal  moments 
by  substituting  the  two  values  of  a'  given  by  that  equation  in  eq. 
(i),  art.  D  II. 

EXAMPLES. 

I.  Determine  the  central  principal  axes  of  the  angle  section  of 
fig.  D  6  and  the  corresponding  moments  of  inertia,  Ix,  ly,  and  Jxy, 
being  45.37,  7.53,  and   —9.67  in.*  respectively. 

Solutions:  (i)  Graphical.  Fig.  D  15  is  the  inertia-circle  for  the 
area  corresponding  to  the  axis  Ox  and  Oy  (constructed  as  explained 
in  the  solution  of  ex.  i,  art.  D  13).  0(i)  and  0(2),  parallel  to  AP^ 
and  AP2,  are  the  principal  axes  at  0,  and  OP^  and  OP^  represent 
(by  the  scale  used)  the  greater  and  lesser  principal  moments  re- 
spectively. 


37S  ,  APPENDIX  D, 

(2)  Algebraic,     Substituting  in  equation  (i),  we  find  that 
tan  2a'  =  2(  -9-67)/(7-53  -45-32)  =0.5118, 
i.e.,  2a' =  27°  6'  or  207°  6';  hence  a'  =  i3°*33'  or  103°  33'. 

Substituting  these  two  values  successively  in  eq.  (i),  art.  D  11, 
we  find  as  the  two  values  of  /« 

/,  =47. 70  in.*     and     7,  =  5.20  in.* 

2.  In  fig.  D  6  (6)  let  a,  b,  and  t  equal  6,  3^,  and  |  in.  respectively; 
then     - 

/x  =  25.32,     /y  =  9.ii,     and     7^^= -11.54  in.* 

Determine  the  central  principal  axes  and  the  corresponding  mo- 
ments of  inertia. 

3.  In  fig.  D6(a)  let  AB=BC=^sm.,  AA' =  CC ^\  m.,  then 
the  centroid  is  0.84  in.  from  AB  and  BC  and  /i  =  /y  =  i.24  in.* 
Determine  the  central  principal  axes  and  the  corresponding  radii 
of  gyration.  Ans.  ^3  =  0.59  in. 

D  15.  Graphical  Determination  of  Moment  of  Inertia  of  a  Plane 
Figure.* — Let  ahha  (fig.  D  16)  be  the  given  figure  whose  mo- 
ment of  inertia  with  respect  to  OX,  say,  is  required,  (i)  Draw 
OjXj  and  O^X^  at  any  convenient  distance  m  from  OX,  and 
through  any  convenient  point  P  draw  a  y  axis,  marking  its  inter- 
section with  O^X^  and  O^X^  M  and  TV  respectively.  (2)  Draw  a 
line  aa  cutting  the  figure  and  parallel  to  the  x  axis,  marking  its 
intersections  with  the  perimeter  a,  and  that  with  the  y  axis  or, 
(3)  Lay  off  Ma'  equal  to  Pa,  and  determine  the  intersections  of 
the  lines  aa'with  OjX,,  marking  them  a'.  (4)  Determine  the  inter- 
sections of  the  lines  Pa'  with  aa,  marking  them  a".  (5)  Repeat  the 
construction  for  other  lines  like  aa,  as  hh,  thus  locating  points  h' . 
(6)  Draw  a  smooth  curve  through  all  points  a",  6",  c",  etc.  (7) 
Measure  the  area  of  the  loops  t  (shaded  in  the  figure) ;  if  A"  denotes 
that  area,  then  the  desired  moment  of  inertia  equals  A'^m^. 

Proof:   The  moment  of  inertia  is  given  by 

i=  fdA'y^=  f     Vwdy, 

w  denoting  any  width  of  the  figure,  as  aa  or  hh.     Let  w'  and  «;" 

*  Logically  this  article  should  appear  in  section  i  of  this  appendix, 
f  If  the  inertia  axis  does  not  cut  the  plane  figure,  then  there  will  be 
only  one  loop. 


SECOND  MOMENTS  OF  y4REAS. 


379 


respectively  denote  the  lengths  a' a'  and  a" a"  (or  6'6'  and  h"h")\ 
then  from  the  geometry  of  the  figtire 

±,y/'uif  =m/w\     and     :ty/w^  =  fn/w, 
the  positive  or  negative  sign  being  taken  according  as  the  ze;'s 


Oj     "/ 


\Sl Ik 


hf 


o 

\ 

p        !    1 

X 

/ 

1      1  \^ 

i 

h  m 

16            \         \l 

V 

=^^.         /A 

\  / 

N        \/             X 

» 

■    5\ 

\ 

/ 

Fig.  D  i6. 

refer  to  widths  above  or  below  the  %  axis.     Miiltiplication  of  the 
equations  gives 

y^/'w'w"  —  m'^/w''w,     or    y'^w=ni^w'\ 
Substituting  this  value  of  y^w  in  the  first  equation  gives 


380  ^  APPENDIX  D. 

EXAMPLE. 

Draw  in  succession  three  circular  arcs  of  30°  each,  3,  2,  and  i  in. 
in  radius  respectively,  and  tangent  to  each  other  so  that  they  form 
approximately  an  elliptical  quadrantal  arc.  Compute  the  moment 
of  inertia  of  the  elliptic  quadrant  with  respect  to  the  major  axis 
of  the  ellipse.  [Suggestion:  Take  the  point  P  at  the  centre  of  the 
ellipse.] 


APPENDIX  E. 

VIRTUAL    WORK. 

E  I.  Definitions.  —  Any  imaginary  displacement  of  a  point  is 
called  a  virtual  displacement  of  that  point.  The  work  which  a 
force  would  do  in  a  real  displacement  of  its  application  point  like 
some  particular  virtual  displacement  is  called  the  virtual  work 
of  that  force  for  that  virtual  displacement. 

The  virtual  work  of  a  force  may  be  computed  by  the  methods 
explained  in  arts.  287  and  288  for  computing  real  work. 

E  2.  Principle  of  Virtual  Work  for  a  Particle. — If  a  particle  is  in 
equilibrium,  then  the  algebraic  sum  of  the  virtual  works  of  all  the 
forces  acting  upon  it  for  any  infinitesimal  virtual  displacement 
equals  zero.* 

Proof:  It  follows  from  art.  292  that  the  algebraic  sum  of  the 
virtual  works  of  the  forces  equals  the  virtual  work  of  their  resultant. 
In  an  infinitesimal  displacement,  the  work  of  each  force  would 
generally  be  an  infinitesimal  of  the  first  order  if  the  displacement 
is  regarded  as  of  that  order;  and  since  the  magnitude  of  the  resultant 
would  not  become  finite  in  such  a  displacement,  the  work  of  the 
resultant  wouM  be  an  infinitesimal  of  the  second  order.  Hence, 
according  to  the  theory  of  infinitesimals,  the  algebraic  sum  of 
the  virtual  works  of  the  forces  acting  on  the  particle  may  be  written 
equal  to  zero. 

Equations  of  virtual  work  for  forces  acting  on  a  particle  in 
equilibrium  are  easily  reduced  to  equations  of  equilibrium  and 
therefore  are  not  specially  useful.  They  are  here  explained  as,  a 
preliminary  to  other  practically  useful  methods. 

EXAMPLES. 

I.  A  bead  A  (fig.  E  i)  is  to  be  supported  on  a  smooth  circular 
wire  (plane  vertical)  in  the  position  shown  by  means  of  a  horizontal 
force.     Compute  the  value  of  the  force. 

*  This  principle  holds  also  for  finite  displacements  if  the  forces  are 
in  equilibrium  at  all  stages  of  the  displacement,  as  is  evident  from  the 
first  statement  in  the  proof  above. 

381 


K 

"y 

1    \ 

1 

\ 

/  _B 

V 

/\ 

A^ 

^ 

^ 

r 

P^ 

38*  -  APPENDIX  E. 

Solution:  The  forces  acting  on  the  bead  are  the  pull  P,  the 
weght  of  the  bead  W,  and  the  reaction  of  the  wire  R.  Assuming 
a  virtual  displacement  from  .4  to  B  (perpendicular  to  R),  the 
forces  remaining  tmchanged  in  magnitude 
and  direction,  then  their  virtual  works  are 
respectively 

PAB  cos  a,  —WAB  sin  a,  and  o. 

The  eqtiation  of  virtual  work  is  therefore 

PAB  cos  a  —  WAB  sin  a  +  0=  o ; 

hence  P  =  W  tan  a. 

'^*      ^'  The    reason    for    choosing    a  virtual  dis- 

placement perpendicular  to  R  should  be  noted.  So  selected,  the 
virtual  work  of  R  is  zero,  and  therefore  R  will  not  appear  in  the 
equation  of  virtual  work,  leaving  but  one  unknown  quantity,  P. 
in  that  equation.  Any  other  direction  for  the  virtual  displacement 
would  introduce  R  into  the  equation  of  virtual  work,  making  two 
tmknowns  in  the  equation;  to  determine  either  unknown  would 
necessitate  the  use  of  a  second  equation  of  virtual  work,  the  vir- 
tual displacement  being  chosen  in  some  direction  not  coincident 
with  the  first. 

It  should  be  noticed  also  that  the  foregoing  equation  of  virtual 
work  after  division  hy  AB  is  an  ordinary  equilibriimi  equation — 
the  one  obtained  by  resolving  the  three  forces  along  AB  and  taking 
the  algebraic  sum  of  the  components. 

2.  Show  by  the  principle  of  virtual  work  that  R=W  sec  a 
in  ex.  I. 

3.  Solve  ex.  5,  art.  129,  by  the  principle  of  virtual  work. 

E  3.  Principle  of  Virtual  Work  for  a  System  of  Particles. — If  every 
particle  of  a  system  is  in  equilibrium,  then  the  algebraic  sum 
of  the  works  of  all  the  forces  (external  and  internal)  acting  on 
the  system  for  any  infinitesimal  virtual  disj^lacement  equals 
zero.  If  the  system  is  a  rigid  one,  then  the  algebraic  stun  of  the 
virtual  works  of  the  external  forces  equals  zero. 

Proof:  According  to  art.  E  2,  the  algebraic  simi  of  the  virtual 
works  of  all  the  forces  acting  upon  any  one  particle  for  any  infini- 
tesimal virtual  displacement  equals  zero;  hence  the  simi  of  the 
virtual  works  of  the  forces  acting  on  all  the  particles  also  equals 
zero. 

If  the  system  is  a  rigid  one,  then  the  algebraic  sura  of  the  virtual 
w^crks  of  all  the  internal  forces  equals  zero  because  the  virtual 


VIRTUAL   IVORK, 


3^3 


work  of  each  pair  O-  internal  forces  constituting  an  action  and 
reaction  equals  zero  (see  art.  293),  and  hence  the  virtual  work  for 
all  pairs,  or  all  the  internal  forces,  equals  zero.  It  follows  there- 
fore that  the  algebraic  sum  of  the  virtual  works  of  all  the  external 
forces  equals  zero. 

In  applying  this  principle,  the  virtual  displacements  should  be 
chosen  so  as  to  make  the  virtual  work  of  the  unknown  forces  not 
desired  equal  to  zero,  thus  eliminating  those  unknown  forces  from 
the  equation  of  virtual  work. 

EXAMPLES. 

I.  A  heavy  bar  rests  against  a  smooth  wall  and  on  a  smooth 
floor  as  shown  in  fig.  E  2,  and  is  prevented  from  slipping  by  a 
horizontal  cord  tied  to  its  lower  end  and  to  the  foot  of  the  wall. 
The  inclination  of  the  bar  being  a,  its  weight  W,  and  its  centre  of 
gravity  at  the  middle,  compute  the  tension 
in  the  cord  by  the  principle  of  virtual 
work. 

Solution :  The  external  forces  acting  on 
the  bar  consist  of  the  pull  of  the  string  P, 
the  weight  of  the  bar  W,  and  the  reactions 
of  the  wall  and  floor,  horizontal  and  ver- 
tical respectively.  If  a  displacement  of 
the  bar  to  the  position  represented  by  the 
dotted  line  is  asstmied,  then  the  virtual 
work  of  each  reaction  will  equal  zero.  If 
the  new  inclination  of  the  bar  be  called 
a-\-da  and  the  length  2/,  then  the  virtual  yig  E  2. 

work  of  the  pull  will  be 

P[2/cosq:  — 2/cos  {a-\-da)]=  —P2I dcosa, 
and  the  virtual  work  of  W  will  be 

—  W[l  sin  (a  +  da)  —  /  sin  a]  =  —  Wl  d  sin  a. 
The  equat'on  of  virtual  work  is  therefore 

—P2I  d  cos  a  —  Wl  ds\Tia  =  o\ 
hence 

P=\W  cot  a. 


2.  Solve  ex.  6,  art.  137,  by  the  principle  of  virtual  work. 

3.  Determine  W  of  ex.  2,  art.  162,  by  the  principle  of  virtual 


work. 


SH  <         APPENDIX  E. 

Solution:  Consider  the  forces  acting  on  the  collection  of  bodies 
(fig.  1446)  consisting  of  the  cross-head,  connecting-rod,  crank- 
axle  and  drum,  hoisting-chain,  and  th»,  suspended  load.  The 
f(?rces  external  to  this  s^^stem  consist  of  P,  the  weights  of  the 
parts,  the  reactions  of  the  cross-head  guides,  and  those  of  the  axle 
bearings.  The  internal  forces  consist  of  the  mutual  pressures  at 
the  pins  A  and  B,  those  between  the  links  of  the  chain,  and  the 
forces  "within"  each  rigid  body.  Now  if  all  rubbing  surfaces 
are  regarded  smooth,  then  the  only  forces  whose  virtual  work  is 
not  zero  for  a  slight  displacement  of  the  cross-head  forward  are 
P  and  the  weights  of  the  parts.  If  the  displacement  of  the  cross- 
head  be  called  ds,  then  that  of  the  suspended  body  is  2. ids,  as 
can  be  shown  by  the  trigonometry  of  the  figure.  Disregarding 
the  weights  of  the  connecting-rod  and  the  chain,  the  equation 
of  virtual  work  becomes 

1000^5  —  1^2.1^5      or     1^=1000/2.1=476  lbs. 

E  4.  Application  of  the  Principle  of  Virtual  Work  to  Statically 
Indeterminate  Problems. — By  a  statically  indeterminate  problem  is 
meant  one  relating  o  a  system  of  forces  in  equilibrium  which  can- 
not be  solved  by  the  principles  of  statics  alone.  Ex.  2,  art.  138, 
is  of  this  class;  the  determination  of  the  reactions  on  a  beam 
supported  at  three  points  is  another.  Trusses  with  superfluous 
or  redundant  members  or  supports  are  also  statically  indeterminate 
(see  art.  F  ii,  Appendix  F).  The  statically  indeterminate  forces 
in  the  cases  just  mentioned  depend  upon  the  way  in  which  the 
beam  or  truss  or  their  supports  deform,  and  the  additional  princi- 
ples needed  relate  to  the  elasticity  of  materials.  For  a  fuller 
treatment  of  this  subject  the  student  is  referred  to  works  on 
"Strength  of  Materials"  and  "Bridges."  The  solution  to  a  simple 
example  only  is  given  here  to  satisfy  the  ciiriosity  of  the  student 
who  has  wondered  over  these  cases. 

EXAMPLES. 

I.  A  square  board  is  suspended  in  a  horizontal  position  by 
means  of  vertical  wires  fastened  at  its  corners;  then  a  heavy  body 
is  p'aced  on  the  board  at  its  centre.  It  is  required  to  determine 
the  tensions  in  the  wires  supposing  that  the  body  weighs  100  lbs., 
and  that  the  wires  fastened  at  A,  B,  C,  and  D  (fig.  E  3)  are  o.oi, 
0.02,  0.03,  and  0.4  sq.  in.  in  cross-section  respectively. 

Solution:    The  forces  acting  on  the  board  are  the  weight  and 


yiRTUAL    WORK. 


385 


the  pulls  of  the  four  wires.     If  any  one  of  these  four  were  known, 
the  other  three  could  be  obtained  from  the  three  conditions  of 


Fig.  E  3. 

equilibrium  for  such  a  system,  parallel  non-coplanar.  Call  the 
pull  first  determined  the  statically  indeterminate  one  and  denote 
it  by  X,  and  the  others  by  P^,  P^,  and  P^  (fig.  E  3a).  Then,  by 
the  conditions  of  equilibrium  (art.  125),  it  is  easy  to  show  that 

P2=X,     and     Pi  =  P3-5oo-X 

Now  instead  of  applying  the  principle  of  virtual  work  to  the 
five  forces  acting  on  the  board,  take  the  force  X  applied  at  D 
and  three  other  vertical  forces  at  A,  B,  and  C,  which  togethe: 
with  X  would  constitute  a  system  in  equilibrium ;  that  system  is 
represented  in  fig.  E  36.  Assume  a  virtual  displacement  like  the 
displacement  which  the  board  actually  underwent  when  the  body 
was  placed  upon  it.  This  displacement  was  permitted  by  the 
elongations  of  the  wires,  and  it  is  now  necessary  to  compute  those 
elongations.  The  elongation  of  a  wire,  not  strained  beyond  its 
** elastic  limit"  (assumed  to  be  the  case  here),  is  given  by  the 
expression  Pl/AE,  in  which  P  denotes  pull,  /  length,  A  area  of 
cross-section,  and  E  Young's  modulus  for  the  material  of  the 
wire.     Hence  the  elongations  are  as  recorded  below: 


Wires. 

Elongations. 

Virtual  Works. 

A 
B 
C 
D 

(5oo-X)//o.oi£: 
XI/0.02E 

(Soo-X)l/o.o3E 
XI/0.04E 

X{soo-X)l/o.oiE 
-XH/0.02E 

X(soo-X)l/o.osE 
-XH/0.04E 

Multiplying  the   forces  by  the  corresponding  virtual  displace- 
ments of  their  application  points  and  affixing  the  proper  signs, 


386  APPENDIX  E. 

we  get  the  expressions  recorded  in  the  last  column  of  the  tabula- 
tion. The  equation  of  virtual  work  after  cancellation  and  reduc- 
tion becomes 

(500—  X)(l/o.OI  + 1/0.03)=  X(l/0.02  + 1/0.04); 

hence        X=40o  lbs.,  and,  according  to  the  foregoing, 
P2  =  ^=400,  P^  =  P^=  100  lbs. 


APPENDIX  F. 

SUPPLEMENT  TO  STATICS. 

Arts.  I  to  1 1  of  this  appendix  relate  to  analysis  of  Roof  Trusses, 
and  arts.  12  to  16  deal  with  some  important  properties  of  the 
funicular  polygon,  useful  in  various  lines  of  applied  mechanics. 

F  I.  Truss  Loads. — Roof-  and  bridge-truss  loads  may  be  classi- 
fied into  permanent,  or  dead,  and  temporary,  or  live.  A  permanent 
or  dead  load  is  one  always  on  the  truss,  while  a  temporary  or 
live  load  is  one  not  always  on  the  truss. 

A  roof -truss  commonly  sustains  dead  loads  only,  as  its  own 
weight  and  that  of  the  roof  covering;  weight  of  snow  and  wind 
pressure  are  live  loads. 

A  bridge  truss  sustains  both  dead  and  live  loads ;  he  first 
consists  of  the  weights  of  the  truss,  floor,  etc.,  and  the  second  of 
the  weights  of  passing  crowds,  cars,  and  wagons,  snow,  wind  pres- 
sure, etc. 

F  2.  Weight  of  Roofing. — The  weight  of  this  can  be  closely 
estimated  for  any  kind  of  covering.  The  following  are  weights  of 
some  roofing  materials  in  pounds  per  square  foot  of  roof  surface. 

Shingling:  tin,  i ;  wood  shingles,  2  to  3 ;  corrugated  iron,  i  to  3 ; 
slate,  8  to  10;   tiles,  10  to  25. 

Sheathing:    Boards,  3  to  5. 

Rafters:    Wood,  1.5  to  3. 

Purlins:   Wood,  i  to  3;  steel,  2  to  4. 

Rafters  and  purlins  are  beams  whose  loads  in  a  given  case  are 
approximately  known,  hence  their  necessary  size  and  weight  can 
be  fairly  accurately  computed. 

F  3.  Weight  of  Trusses. — The  actual  weight  of  a  truss  can  be 
determined  only  after  it  is  designed.  Its  probable  weight  must 
be  known  for  the  analysis,  and  this  is  estimated  from  the  weights 
of  similar  existing  trusses  or  computed  from  a  formula  derived  from 
the  actual  weights  of  existing  trusses.  The  following  is  such  a 
formula  for  the  weights  of  steel  trusses:* 

W  =  al(i  +  1/2S), 

*  From  "  Modem  Framed  Structures,"  Johnson,  Bryan,  and  Tumeaure. 

387 


388  ,        APPENDIX  F, 

W  denoting  the  weight  of  a  truss  in  pounds,  a  the  distance  between 
adjacent  trusses,  and  /  the  span,  both  in  feet. 

The  probable  weight  of  a  wooden  trus^  is  about  three-fourths 
of  that  given  by  the  above  formula. 

F  4.  Weight  of  Snow. — The  probable  weight  of  snow  which 
may  have  to  be  borne  by  a  roof-  russ  depends,  of  course,  on  loca- 
tion. In  that  part  of  the  United  States  where  it  is  necessary  to 
allow  for  a  snow  load  the  assumed  weight  varies  from  10  to  30  lbs. 
per  sq.  ft.  of  area  covered  b}^  the  roof. 

F  5.  Wind  Pressure. — The  intensity  of  the  pressure  of  a  wind 
blowing  normally  against  a  plane  surface  is  approximately  propor- 
tional to  the  square  of  the  velocity  of  the  wind;  it  may  be  com- 
puted from  the  formula 

p  =  o.oosv^, 

p  denoting  intensity  in  Ibs./ft.^,  and  v  velocity  in  mi./hr. 

An  intensity  of  30  lbs./ ft. ^  corresponds  by  this  formula  to  a 
velocity  of  77.5  mi./hr. 

The  intensity  of  the  pressure  of  a  wind  blowing  obliquely 
against  a  surface  depends  on  the  inclination  as  well  as  the  velocity. 
The  following  formula  for  intensity  on  plane  surfaces  is  generally 
regarded  as  reliable: 

pi  =  p  sin  i/  (i  +sin^  i) ; 

«  denoting  inclination  of  the  surface  to  the  wind, 
pi  intensity  on  the  inclined  surface,  and 
p  intensity  on  a  surface  normal  to  the  wind. 

For  an  intensity  on  a  normal  surface  (p)  of  30  lbs.  per  sq.  ft., 
the  intensities  on  oblique  surfaces  according  to  the  foregoing 
formula  are  as  follows: 

i=io,     20,     3.0,     40,     50,     60-90  degrees 
pi=  10,     18,     24,     27,     29,     30  lbs./ ft. ^ 

The  direction  of  the  pressure  of  a  wind  blowing  obliquely 
against  an  inclined  plane  surface  is  practically  normal  to  the 
surface.  In  computing  wind  pressures  on  roofs,  the  wind  is  sup- 
posed to  blow  horizontally  and  at  right  angles  to  the  axis  of  the 
building. 

F  6.  Computation  of  "Apex  Loads."  —  The  weight  of  the  roof 
covering,  including  rafters  and  purlins,  comes  upon  the  trusses 
at  the  points  where  they  support  the  purlins ;  likewise  the  pressure 
due  to  wind  and  snow.     Sometimes  all  the  purlins  are  supported 


SUPPLEMENT  TO  STATICS. 


389 


at  the  joints  of  the  trusses:  in  such  cases  the  loads  mentioned 
act  upon  the  trusses  at  their  joints.  However,  the  roof,  snow,  and 
wind  loads  are  always  assumed  to  be  applied  to  the  truss  at  its 
upper  joints.  This  assumption  is  equivalent  to  neglecting  the 
bending  effect  due  to  the  pressure  of  those  purlins  which  are  not 
supported  at  joints ;  this  effect  can  be  computed  separately. 

The  weight  of  the  truss  itself  is  assumed  to  come  upon  the  truss 
at  its  upper  joints;  this,  of  course,  is  not  exactly  correct.  Most 
of  the  weight  does  come  upon  the  upper  joints,  for  the  upper  mem- 
bers are  much  heavier  than  the  lower,  and  the  assumption  is 
sufficiently  correct  in  most  cases. 

EXAMPLE. 

It  is  required  to  compute  the  apex  loads  for  the  truss  of  fig.  F  i 
due  to  dead  load,  snow,  and  wind.  Assume  that  the  truss  is  a 
steel  one,  total  roofing  weighs  15  lbs./ ft. ^,  snow  weight  is  10  Ibs./ft.^, 
horizontal,  and  the  normal  wind  pressure  is  30  Vos./it? 


Solution:   The  formula  (art.  F  3)  gives  for  the  probable  weight 
of  truss  (see  fig.  F  i  for  dimensions) 

13X42(1  +  42/25),     or     1463  lbs. 

The  area  of  roofing  sustained  by  one  truss  is  about  48^X13,  or 
630^  ft.^;  hence  the  weight  of  it  is  630^X15,  or  9457  lbs.  The 
permanent  or  dead  load  therefore  equals 

1463+9457,     or     11,920  lbs. 

The  area  covered  by  the   roofing  supported  by  one  truss  is 


39®  v  APPENDIX  F. 

42  X13,  or  546  ft.^;  hence  the  snow  load  borne  by  one  truss  equals 
546X10,  or  5460  lbs. 

The  angle  which  the  wind,  directed  as  assumed  in  art.  F  5, 
makes  with  the  roof  is  30°;  hence  the  intensity  of  wind  pressure 
on  the  roof  is  24  lbs./ ft. ^  (see  art.  F  5),  and  the  total  wind  pressure 
borne  by  one  truss  is 

(24^X13)24,     or     7566  lbs. 

The  dead  load  is  proportioned  among  the  five  upper  joints, 
but  :oints  (i)  and  (5)  sustain  only  one-half  as  much  as  the  others. 
Hence  for  joints  (i)  and  (5)  the  apex  load  is  one-eighth,  and 
for  joints  (2),  (3),*  and  (4)  one-fourth  of  the  total  load. 

The  snow  load  is  proportioned  among  the  joints  just  like  the 
dead,  i.e.  one-eighth  at  joints  (i)  and  (5)  and  one-fourth  at 
joints  (2),  (3),  and  (4). 

The  wind  load  coming  upon  one-half  of  the  roof  only,  the  left 
half,  say,  is  proportioned  among  joints  (i),  (2),  and  (3);  one-fourth 
at  joints  (i)  and  (3),  and  one-half  at  joint  (2). 

F  7.  Determination  of  Reactions.  —  To  analyze  a  truss,  it  is 
generally  necessary  to  determine  the  reactions  due  to  the  dead 
and  wind  loads  separately,  and  sometimes  those  due  to  the  snow 
load.  Each  ofthese  load  systems  together  with  the  reactions 
due  to  it  consmutes  a  system  of  forces  in  equilibrium,  and  the 
unknowns  in  the  system  (the  reactions)  can  be  determined  by 
the  principles  of  statics  in  all  ordinary  cases.  The  determination 
usually  falls  under  arts.  139,  140,  or  141.  Two  cases,  however 
(arts.  F  8  and  F  9),  need  further  explanation. 

F  8.  Reactions  on  a  Rigid  Truss  due  to  Wind  Pressure. — (a)  If 
one  end  of  the  truss  rests  on  rollers  *  and  one  end  is  fixed,  the 
reactions  are  statically  determinate.  The  one  at  the  roller  end 
is  practically  vertical,  the  direction  of  the  other  is  unknown  at 
the  outset;  the  reactions  may  be  completely  determined  by 
methods  of  art.  140. 

(6)  If  both  ends  of  the  truss  are  fixed,  the  reactions  are  statically 
indeterminate,  for  the  magnitude  and  direction  of  each  are  unknown 
and  there  are  but  three  conditions  of  equilibrium  available.  Usu- 
ally one  of  the  following  assumptions  is  made  to  determine  the 
reactions : 

*  Rollers  are  usually  placed  under  one  end  of  a  long  truss  to  allow 
it  to  expand  and  contract  freely. 


SUPPLEMENT   TO  STATICS. 


391 


(i)  They  are  parallel  to.  the  resultant  wind  pressure  on  the  roof. 
(2)  Their  horizontal  components  each  equal  one-half  of  the  total 
horizontal  wind  pressure,  or,  otherwise   stated,  each  support 
takes  one-half  of  the  horizontal  thrust  of  the  wind. 

Once  the  first  assumption  is  made,  the  problem  of  determining 
the  reactions  is  reduced  to  that  of  art.  139.  When  the  second 
assumption  is  made,  the  total  horizontal  wind  pressiu-e  or  thinist 
should  be  computed  first — easily  done  from  the  force  polygon 
for  the  wind  loads;  then  imagine  each  reaction  resolved  into  its 
horizontal  and  vertical  components,  and  noting  that  the  horizontal 
components  equal  one-half  of  the  thrust,  there  remain  but  two 
unknown  forces  in  the  wind  system,  namely,  the  vertical  compo- 
nents of  the  reactions.  Their  determination  also  falls  under 
art.  139. 

Obviously  the  first  of  these  two  assumptions  is  wrong  if  the 
resultant  wind  pressure  is  horizontal  and  does  not  pass  through 
the  supports,  for  then  the  two  reactions  (both  horizontal)  could 


IXIXIXIXIXIXM 


Fig.  F  2. 

not  possibly  balance  the  resultant.  In  such  cases  and  in  approxi- 
mations thereto  the  second  assumption  is  made;  examples  are 
a  "bent"  of  a  mill  building  and  a  framed  tower  (fig.  F  2). 


Fig.  F  3. 

F  9.  Reactions  on  a  Three-hinged  Arch.  —  A  three-hinged  arch 
consists  of  two  rigid  trusses  pinned  together  and  each  to  a  support ; 
see  fig.  F  3,  pins  at  a,  b,  and  c.     The  reactions  at  the  supports  are 


392  ._  APPENDIX  F. 

unknown  at  the  outset,  in  magnitude  and  direction,  even  for 
vertical  loads.  At  first  thought  this  appears  to  be  like  case  (6)  of 
the  preceding  article,  but  the  third  hijige  really  makes  the  reac- 
tions statically  determinate. 

There  are  many  solutions  of  this  problem — two  are  here  given; 
each  may  be  carried  out  graphically  or  algebraically,  but  they 
are  respectively  adapted  to  graphic  and  algebraic  methods. 

(i)  Determine  the  reactions  due  to  the  loads  on  each  half  of 
the  arch  separately,  and  then  combine  these  partial  reactions. 
Let  A  and  B  denote  the  total  reactions  at  a  and  h  respectively, 
Ar  and  Br  the  reactions  at  those  points  due  to  the  loads  on  {R), 
and  Ai  and  Bi  those  due  to  loads  on  (L). 

Supposing  first  that  only  {R)  is  loaded,  the  part  (L)  is  under 
the  action  of  two  forces  only,  Ar  and  the  pressure  at  c,  these  two 
being  in  equilibrium  must  be  collincar,  and  hence  they  act  along  ac. 
The  external  forces  on  the  entire  arch  now  (loads  on  {R)  only) 
are  all  known  except  two,  Ar  and  Br,  but  the  action  line  of  ^r 
and  the  application  point  of  Br  are  known ;  these  unknown  forces 
may  hence  be  determined  by  methods  of  art.  140. 

Similarly  it  can  be  shown  that  Bi  acts  along  he,  and  hence  Bi 
and  A I  may  be  determined  like  Br  and  Ar.  Finally  >  composition 
of  ^r  and  Ai  and  of  Br  and  Bi  gives  A  and  B 

The  pressures  exerted  by  {R)  and  (L)  upon  each  other  can  be 
easily  obtained  from  the  component  .reactions  at  a  and  h.  The 
pressure  which  {R)  exerts  on  (L)  is  identical  with  the  resultant 
oi  Ar  reversed  and  Bl,  and  the  pressure  which  (L)  exerts  on  {R) 
is  identical  with  the  resultant  of  Bi  reversed  and  Ar',  these  pres- 
sures are  of  course  equal,  opposite,  and  col  linear. 

(2)  Imagine  the  total  reactions  at  a  and  h  resolved  into  two 
components  one  of  which  acts  along  the  line  ah ;  these  components 
may  readily  be  computed  and  then  compounded  to  get  the  actual 
reactions.  Thus,  call  the  components  acting  along  ah,  A^  and  5^ 
respectively,  and  the  others  Ay  and  By.  "Taking  moments"  of 
all  the  forces  on  the  arch  with  origin  at  a  and  h  gives  Ay  and  By. 
Taking  moments  for  all  the  forces  on  the  part  {R)  with  origin 
at  c  gives  B ^,  and  for  all  the  forces  on  (L)  with  origin  at  c  gives  A^. 
If  the  loads  are  vertical,  and  Ay  and  By  are  taken  vertical,  then 
A^  and  B  ^  are  equal  and  opposite. 

The  pressures  at  c  can  now  be  readily  determined  by  considera- 
tion of  all  the  forces  acting  on  either  {R)  or  (L). 

F  10.   Maximum   Stresses.  —  The    complete    analysis    of    a   truss 


SUPPLEMENT   TO  STATICS. 


393 


should  include  a  determination  of  the  stress  in  each  member  due 
to  (a)  the  dead  load,  {b)  the  snow  load,  {c)  wind  pressure  right,  and 
{d)  wind  pressure  left. 

When  all  the  snow  apex  loads  are  the  same  fractional  part  of 
the  dead  apex  loads,  the  "snow  stress'  in  any  member  equals 
that  same  fractional  part  of  the  permanent  stress  in  that  member; 
no  stress  diagram  for  snow  loads  is  therefore  needed,  but  the  snow 
stresses  are  readily  computed  from  the  permanent  stresses,  con- 
veniently by  slide-rule. 

If  the  truss  is  symmetrical  and  fastened  at  both  ends,  then 
the  stress  in  any  member  when  the  wind  blows  from  the  left  is 
just  like  that  in  the  symmetrical  member  when  the  wind  blows 
from  the  right;  only  one  stress  diagram  is  therefore  necessary, 
"wind  left  stresses"  being  obtainable  from  a  "wind  right  stress 
diagram."  But  when  one  end  of  the  truss  rests  on  rollers  and 
one  end  is  fixed,  then  separate  analyses  for  wind  right  and  left  are 
necessary. 

A  record  of  the  stresses  should  be  made  as  fast  as  they  are 
determined,  the  kind  of  stress  being  indicated  as  well  as  the  amount. 
The  following  is  a  convenient  form: 


Stress. 

Member. 

D.  L. 

S.  L. 

W.  R. 

W.  L. 

Maximum. 

ab 
be 

+  14,000 
—  3,200 

+  7,000 
—  1,600 

+  10,600 
+  4,000 

+  1,400 
—  1,000 

+  31.600 
—  5,800,  +800 

The  abbreviations  are:    D.  L.,  dead  load;    S.  L.,  snow  load; 
wind  right;  W.  L.,  wind  left;  +,  tension;   — ,  compression. 


W.   R., 


By  maximum  stress  for  any  member  is  meant  the  greatest 
stress,  tension  or  compression,  to  which  it  may  be  subjected, 
due  to  any  possible  combination  of  loads  which  may  come  upon 
the  truss.     The  possible  combinations  of  the  usual  loads  are: 


I.  Dead. 

3.  Dead  and  wind  right. 

5.  Dead,  snow,  and  wind  right. 
For  member  ab  the  combination 
duces  the  greatest  stress,  31,600  lbs.,  tension.  No  combination 
produces  a  compression  in  that  member.  For  member  be,  the  com- 
bination of  D.L.,  S.L.,  and  W.L.  produces  the  greatest  compression, 


2.  Dead  and  snow. 

4.  Dead  and  wind  left. 

6.  Dead,  snow,  and  wind  left. 

of  D.L.,   S.L.,   and  W.R.  pro- 


394 


APPENDIX  R 


5800  lbs.,  and  the  combination  of  D.L.  and  W.R.  produces  the 
greatest  tension,  800  lbs. 

Some  assume  that  great  snow  and  wind  loads  will  not  come 
upon  the  truss  at  the  same  time,  and  they  exclude  combinations 
5  and  6  in  a  computation  for  maximum  stresses. 

Fii.  A  Classification  of  Frames. — A  frame  may  be  either  (a) 
complete,  (b)  incomplete,  or  (c)  redundant.  In  the  following  defini- 
tions, it  is  assumed  that  the  frame  is  pin-connected ;  a  riveted  frame 
is  classified  as  though  it  were  pin-connected. 

(a)  A  complete  frame  is  one  which  would  not  maintain  its 
shape  under  all  conditions  of  loading  with  any  member  removed, 
i.e.,  it  is  indeformable  and  without  superfluous  or  redundant 
members.     See  fig.  F  4(a)  for  an  example.     Trusses  of  this  class 


C6) 


(c) 


Fig.  F  4. 

are  statically  determinate. 

(6)  An  incomplete  frame  is  one  which  might  maintain  its 
shape  under  certain  conditions  of  loads,  but  to  do  so  for  all  condi- 
tions would  require  one  or  more  additional  members,  i.e.,  it  is 
generally  deformable.  See  fig.  F  4(6)  for  an  example.  This  truss 
would  maintain  its  shape  for  symmetrical  loading,  but  not  for 
any  other  distribution.  Incomplete  frames  loaded  so  that  they 
will  maintain  their  shape  are  s'atically  determinate. 

(c)  A  redundant  frame  is  one  which  maintains  its  shape  under 
all  conditions  of  loads  and  would  do  so  with  fewer  members,  i.e., 
it  is  indeformable,  but  has  superfluous  or  redundant  members.* 
See  fig.  F  4{c)  for  an  example:  removal  of  either  diagonal  of  the 
rectangle  would  still  enable  the  truss  to  sustain  any  load.  Trusses 
of  this  kind  are  always  statically  indeterminate. 

Criteria  for  Classification  of  a  Frame. — Let  m  denote  the  num 
ber  of  members  in  the  frame  and  /  the  number  of  joints ;  then  for 
a  frame  which  is 

complete,  w  =  2/  —  3 ; 
inc  mplete,  w<2y— 3; 
redundant,     m>2J  —  ^. 


*  These  members  are,  or  ought  to  be,  structurally  useful,  but  are  so 
named  because  the  frame  is  indeformable  without  them. 


SUPPLEMENT   TO  STATICS,  395 

These  criteria  assume  (see  proof  below)  that  every  member  of 
the  truss  can  sustain  bo  h  tension  or  compression.  If  the  truss 
has  members  which  can  sustain  only  tension  or  compression  and 
if  any  one  of  those  members  has  a  counter,*  then  both  the  main 
member  and  its  counter  should  not  be  included  in  ni. 

Proo  :  Evidently  the  simplest  complete  frame  is  a  triangle; 
it  has  three  joints  and  three  members.  It  can  be  extended  so  as 
to  remain  complete  only  by  the  addition  of  two  members  for 
each  new  joint;  thus,  if  m'  and  ;'  denote  the  number  of  new  mem- 
bers and  joints  respectively, 

m'  =  2/'. 

Also,  m'+  3  =  2;'+  3  =  2(y'+  Z)-^, 

or  m=2]  —  ^. 

This  being  the  relation  for  a  complete  frame,  those  already  given 
for  incomplete  and  redundant  frames  follow. 

F  12.  Moments  of  Forces  Determined  Graphically. — Let  ab  (fig. 
F  5)  be  the  action  line  of  a  force,  AB  the  magnitude  and 
direction,  and  P  a  moment  origin.     From  any  convenient  pole  O 


1  ini=xit.  1  in=y1bB, 

Fig.  F  5. 

draw  the  rays  OA  and  OB,  and  then  from  any  point  on  ab  the  corre- 
sponding  strings   oa   and   ob.      Next   draw   a  line   parallel   to   ab 

*  A  counter  is  a  member  whose  main  function  is  to  relieve  another 
member  from  a  stress  of  a  kind  for  which  the  latter  was  not  designed. 
Thus,  if  both  diagonal  members  in  the  quadrilateral  of  fig.  ¥4(0)  con- 
sist of  rods,  neither  can  sustain  a  compression,  and  any  load  which  tends 
to  put  a  compression  on  one  will  put  tension  on  the  other;  hence  either 
member  may  be  regarded  as  a  counter  to  the  other,  but  the  one  which 
would  be  stressed  under  the  permanent  load  would  be  called  the  main 
member,  and  the  other  the  counter. 


396 


APPENDIX  F. 


through  P,  and  note  its  intersections  M  and  N  with  the  strings 
oa  and  oh.  Then  measure  the  intercept  M.N  by  the  scale  of  the 
space  diagram  and  the  perpendicular*  OC  from  the  pole  to  AB 
by  the  scale  of  the  vector  diagram;  so  measured,  their  product 
equals  the  moment  of  the  force.  For,  by  definition,  the  moment 
of  the  force  equals 

(the  force)  X  (the   Sirm)  =  {AB-y){PQ-x), 

and  from  the  similar  triangles  OAB  and  LMN, 

AB/OC  =  MN/PQ,     or     {MN){OC)  =  iAB){PQ) ; 
hence 

(MN-x){OC-y)  =  iAB-y){PQ-x). 

OC  is  called  the  pole  distance  of  the  force  AB,  and  if  we  call  MN 
the  intercept  of  the  force  with  respect  to  P,  then  we  may  state 
that  the  moment  of  a  force  with  respect  to  a  point  equals  the  prod- 
uct of  its  intercept  with  respect  to  that  point  and  its  pole  dis- 
tance, it  being  understood  that  the  intercept  and  pole  distance 
are  properly  scaled  as  explained. 

F  13.  Sum  of  the  Moments  of  any  Number  of  Forces. — This  is 
most  conveniently  computed  graphically  by  determining  its  equal, 
the  moment  of  their  resultant.  Thus,  to  find  the  sum  of  the 
moments  of  the  four  forces  represented  in  fig.  F  6(a)  with  respect 


Tig.  F  6. 

to  P,  first  determine  the  resultant  {AE  and  ae)  and  then  the  **  inter- 
cept" {MN)  and  the  "pole  distance"  {OK)  of  the  resultant. 
The  product  of  the  intercept  and  pole  distance  is  the  moment 
sought.  Notice  that  the  intercept  is  that  cut  off  on  the  line  through 
the  moment  origin  parallel  to  the  resultant  by  the  strings  **  corre- 
sponding" to  the  resultant.  It  is  unnecessary  to  actually  draw 
the  action  line  of  the  resultant,  but  its  approximate  position  and 


SUPPLEMENT  TO  STATICS. 


397 


the  sense  of  the  resultant  must  be  known  in  order  to  fix  the  sign 
of  the  moment,  here  negative. 

When  once  a  force  and  a  funicular  polygon  for  any  system 
of  forces  have  been  constructed,  then  the  algebraic  sum  of  the 
moments  of  any  number  of  them  which  are  represented  consecu- 
tively in  the  force  polygon  can  be  computed  as  just  explained. 

F  14.  Parallel  Forces  in  Equilibrium. — The  algebraic  sum  of 
the  moments  of  all  the  forces  on  either  side  of  any  moment  origin 
can  be  easily  read  from  any  funicular  polygon  of  the  system  pro- 
vided that  these  same  forces  occur  consecutively  in  the  corre- 
sponding force  polygon.* 

Thus  let  the  loads  and  reactions  on  the  beam  represented  in 
fig.  F  7  be  the  system  considered;    ABCDEA  is  a  force  polygon 


Fig.  F  7. 

satisfying  the  requirement  specified,  and  oa,  oh,  oc,  od,  and  oe  a 
funicular  polygon.  The  resultant  of  the  forces  to  the  left  of  P^  is 
EB,  and  according  to  the  preceding  article  the  algebraic  sum  of 
the  moments  of  those  two  forces  with  respect  to  P  equals  the 
intercept  cut  off  by  the  strings  oe  and  ob  on  the  vertical  line  through 
P  multiplied  by  the  pole  distance  OK. 

Since  this  pole  distande  is  the  same  for  all  resultants  of  groups 
of  the  forces,  the  algebraic  sums  of  the  moments  of  the  forces  to 
the  left  or  right  of  different  moment  origins  are  proportional  to 
the  intercepts,  or  ordinates,  in  the  funicular  polygon  immediately 
below  the   origins;    the   figtire  formed  by  the   funicular  polygon 

*  The  property  of  the  funicular  polygon  here  being  described  is 
utilized  mostly  in  connection  with  beam  problems — loads  and  reactions 
vertical.  If  the  loads  and  reactions  are  represented  in  the  force  polygon 
in  the  order  in  which  their  application  points  occur  around  the  beam, 
then  the  forces  appear  consecutively  as  above  described. 


398 


APPENDIX  F. 


is  therefore  called  a  "moment  diagram."  If  the  scale  of  the 
space  diagram  is  i  in.  =  ic  ft.  and  the  pbie  distance  scales  z  lbs., 
then  the  scale  of  the  moment  diagram  is  i  m.  =  xz  ft. -lbs.  Select- 
ing the  scale  and  the  pole  so  as  to  make  x  and  z  "round  numbers," 
then  the  scale  of  the  moment  diagram  will  be  simple,  and  by  it 
moments  may  be  read  directly  from  the  diagram. 


EXAMPLES. 

1.  Compute  graphically  the  moment  of  each  force  of  ex.  i, 
art.  45,  and  also  the  moment  of  their  resultant.  Compare  the 
last  with  the  algebraic  sum  of  the  moments  of  the  forces. 

2.  A  beam  20  ft.  long  rests  on  end  supports  and  sustains  loads 
of  8000,  6000,  and  10,000  lbs.  2,  8,  and  18  ft.  respectively  from 
the  left  end.  Construct  a  moment  diagram  for  all  the  forces 
acting  on  the  beam;  state  its  scale  and  give  the  algebraic  sum 
of  the  moments  of  all  the  forces  on  the  left  half  with  respect  to  the 
middle. 

3.  Solve  ex.  2  supposing  that  the  supports  are  3  ft.  from  the 
ends. 

F  15.  To  "  Pass "  a  Funicular  Polygon  through  Three  Points. — 
By  this  is  meant  the  construction  of  a  funicular  polygon  for  a 
given  system  of  forces  so  that  three  specified  strings  shall  pass 
through  three  given  points,  a  construction  often  made,  especially 
in  the  design  of  a  masonry  arch. 

It  will  first  be  shown  how  to  pass  a  funicular  polygon  through 
two  points.     Let  ab,  be,  cd,  and  de  (fig.  F  8)  be  four  forces;    re- 


FiG.  F  8. 


quired  to  construct  a  funicular  polygon  so  that  the  strings  oa  and 
od  shall  pass  through  p^  and  p^  respectively.  Imagine  the  forces 
included  within  the  rays  OA  and  OD  in  the  force  polygon  to  act 


SUPPLEMENT  TO  STATICS.  399 

upon  a  beam  supported  at  p^  and  p^,  the  supports  being  such 
that  the  reactions  are  statically  determinate.  For  simplicity 
imagine  that  the  supports  are  such  that  the  reactions  are  parallel 
to  each  other  and  therefore  to  the  resultant  of  the  loads  applied  to 
the  beam;  then  determine  the  magnitudes  of  those  reactions. 
Thus,  having  drawn  the  force  polygon,  join  A  and  D  to  get  the  direc- 
tion of  the  reactions;  then  draw  their  action  lines,  parallel  to  AD. 
Next  construct  a  funicular  polygon  and  note  the  intersections 
of  the  first  and  last  strings,  oa  and  od,  with  the  action  lines  of 
the  reactions  through  p^  and  p^  respectively.  Join  these  intersec- 
tions— the  line  is  the  closing  string  (see  art.  139) — and  draw  a 
ray  parallel  to  that  line,  noting  its  intersection  Q  with  AD;  then 
QA  and  DQ  denote  the  reactions  at  p^  and  p^  respectively.  Next 
choose  a  new  pole  O'  anywhere  on  a  line  through  Q  parallel  to  p^p^, 
and  construct  a  corresponding  funicular  polygon  beginning  at 
P\  or  p2 ;   this  new  polygon  will  pass  through  both  points. 

Proof:  The  three  loads  AB,  BC,  and  CD  and  the  reactions  are  in 
equ  librium,  and  every  funicular  polygon  for  those  forces  must  close. 
Hence  if  p^p2  is  t  aken  as  the  string  o'q  of  a  new  funicular  polygon 
(and  hence  a  new  pole  O'  on  the  line  through  Q  parallel  to  pip-^ 
and  then  the  strings  o'a,  o'h,  etc.,  be  drawn,  o'd  must  pass  through 
/?2  to  close  the  polygon. 

To  pass  a  funicular  polygon  through  three  points:  By  the 
method  explained  in  the  foregoing,  determine  a  ray  Q'O',  the  funicu- 
lar polygon  corresponding  to  which  will  pass  through  any  pair  of 
the  three  points  as  prescribed ;  then  determine  another  ray  Q"0'\ 
the  funicular  polygon  corresponding  to  which  will  pass  through 
another  pair  of  the  three  points  as  prescribed.  Then  with  the  inter- 
section of  Q'O'  and  Q"0"  as  a  pole,  construct  a  funicular  polygon 
drawing  one  of  ^  the  specified  strings  through  its  specified  point ; 
it  will  be  found  that  the  polygon  passes  as  required  through  the 
other  two  points. 

As  an  illustration,  take  five  parallel  forces,  ah,  be,  ed,  de,  and  df 
(fig.  F  9),  the  requirement  being  to  make  oa,  od,  and  of  pass  through 
p^,  p2,  and  p3  respectively.* 

*  This  is  the  form  in  which  the  practical  problem  appears,  i.e.,  the 
forces  are  parallel,  two  of  the  points  embrace  all  the  forces  and  the 
third  is  somewhere  between ;  the  first  and  last  strings  are  to  pass  through 
the  first  two  points,  and  that  string  is  to  pass  through  the  third  point 
which  extends  between  the  action  lines  of  the  two  forces  adjacent  to 
that  point. 


40O 


y4PPENDlX  F. 


The  first,  or  preliminary  funicular  polygon,  is  the  upper  one, 
corresponding  to  the  pole  O^]    its  strings  o^a,    o^d,  and  oj  with 


Fig.  F  9. 

the  verticals  through  p^,  p^,  and  p^  respectively  determine  the 
closing  strings  o^q'  and  o^q".  Lines  parallel  to  these  through  O^ 
determine  Q'  and  Q",  and  lines  through  these  points  parallel  to 
p^p2  and  p2pz  respectively  determine  the  pole  O,  the  funicular 
polygon  corresponding  to  which  if  started  at  p^,  p^,  or  p^,  as  pre- 
scribed, will  pass  through  the  other  two  points. 

Before  beginning  the  construction  of  the  final  funicular  polygon, 
it  may  be  well  to  check  the  location  of  the  pole  0  by  means  of  the 
third  closing  line  13 ;  this  line  serves  to  locate  Q"'  {FQ'"  and  Q"'A 
being  reactions  on  the  imaginary  beam  supported  at  p-^  and  p^ 
and  sustaining  all  the  forces  as  loads) ;  a  line  through  Q'"  parallel 
to  p^p^  will  pass  through  O  if  no  error  has  been  made. 


EXAMPLE. 


Assume  completely  six  parallel  forces  and  select  three  points 
in  the  space  diagram  not  in  the  same  straight  line,  and  then  con- 
struct a  funicular  polygon  so  tljat  three  of  its  strings  will  pass 
throuj^h  the  points. 


SUPPLEMENT   TO  STATICS. 


401 


F 16.  Relation  between  Two  Funicular  Polygons  for  a  Given  Force 
System  Drawn  from  Different  Poles. — The  intersections  of  correspond- 
ing strings  of  such  polygons  lie  on  a  straight  line  parallel  to  the 
line  joining  the  two  poles. 

Proof:    Let  ah,  be,  and  cd  (fig.  F  10)  be  three  forces,  oa,  oh,  oc. 


Fig.  F  id. 

and  od  being  their  funicular  polygon  for  the  pole  O,  and  o'a,  o'h, 
o'c,  and  o'd  their  funicular  polygon  for  the  pole  0' ',  £he  intersec- 
tions of  corresponding  strings  of  these  polygons  are  m^,  m^,  w,, 
and  m^. 

The  forces  BO  and  OC  are  equivalent  to  BC;  so  also  are  BO' 
and  O'C*  Therefore  BO,  OC,  O'B,  and  CO'  are  in  equilibrium, 
and  the  resultant  of  any  pair  of  these  four  balances  the  resultant 
of  the  other  pair.  Now  the  resultant  of  O'B  and  BO  acts  through 
Wj,  and  thfe  resultant  of  OC  and  CO'  acts  througn  m^',  since  these 
resultants  balance,  they  must  coincide,  i.e.,  each  acts  through 
w,  and  W3.  The  direction  of  the  resultant  of  the  first  pair  is  O'O 
(and  that  of  the  second  is  00'),  and  since  the  action  line  of  a  force 
is  parallel  to  its  direction,  mjWg  is  parallel  to  00'. 

In  a  similar  manner  it  might  be  shown  that  m^m^  and  m^m^  are 
parallel  to  00' ;  thus  the  proposition  is  proved. 

By  means  of  the  property  discussed  in  the  foregoing,  a  string 
may  readily  be  drawn  to  an  inaccessible  intersection,  e.g.,  one 
which  falls  beyond  the  limits  of  a  drawing.  Thus,  let  ah,  he,  and 
cd  (fig.  F  11)  be  three  forces,  oa  and  oh  two  strings  of  a  ^unicular 


*  The  sequence  of  these  letters  indicates  the  sense  of  the  force  desig- 
nated. 


402 


APPENDIX  F. 


polygon  which  it  is  desired  to  complete.     The  string  oc  ordinarily 
would  be  drawn  parallel  to  OC  through  the  intersection  of  ob  and  be 


Fig.  F  II. 

here  beyond  the  limits  of  the  drawing.*  Choose  a  new  pole  O' ;  from 
any  point  n  in  be,  draw  the  strings  o'b  and  o^c,  and  note  the  inter- 
section m'  of  o'b  and  ob ;  draw  through  ni'  a  parallel  to  OO^,  and 
note  its  intersection  w"  with  oV;  finally  draw  through  w"  a  line 
parallel  to  OC.     This  is  the  desired  string  oc. 

Notice  that  it  will  always  be  possible  to  choose  O'  and  n  so 
that  the  points  m'  and  w"  will  fall  upon  the  drawing. 

F  17.  To  Close  a  Gauche  f  Polygon  with  Three  Sides  whose  Directions 
are  Given. — This  is  a  geometrical  construction  used  in  the  graphical 
solution  of  the  following  problem:  A  system  of  concurrent  non- 
coplanar  forces  is  in  equilibrium  and  all  the  forces  are  known 
except  three,  whose  action  lines  only  are  known;  required  to 
determine  their  magnitude  and  senses. 

Let  there  be  four  forces  in  the  system,  F  being  the  wholly 
known  force  and  F^,  F^,  and  F3  the  partially  known  ones,  and 
let  their  action  lines  be  as  represented  in  fig.  F  1 2  (a)  .t   The  polygon 

*  In  such  a  case  a  draughtsman  usually  tacks  a  sheet  of  paper  to  his 
board  which  will  take  the  intersection  desired;  often,  however,  such 
an  intersection  is  very  acute — therefore  indefinite — and  this  special 
construction  is  advantageous. 

t  A  gauche  polygon  is  one  whose  sides  are  non-coplanar. 

X  See  art.  49  for  notation  and  scheme  of  representing  concurrent 
non-coplanar  forces. 


SUPPLEMENT   TO  STATICS. 


403 


for  the  system  closes,  so  let  it  be  called  ABCDA,  AB,  BC,  CD, 
and  DA  representing  the  magnitudes  and  directions  of  F,  F^,  Fj, 


Fig.  F  12. 


and  F,  respectively,  and  let  A'B'C'D'A'  and  A"B''C"D''A''  be 
its  horizontal  and  vertical  projections — their  construction  will 
be  explained  presently. 

The  point  J'  (of  any  polygon  such  as  c'c"d"d\  similar  to 
C'C"D"D'),  the  point  Q  (construction  for  which  is  apparent), 
and  the  point  D'  are  in  the  same  straight  line,  for  regard  A"D" , 
B"C'\  and  B'C  (indefinite  in  length)  as  the  action  lines  of  three 
forces  such  that  D'D"C"C'D'  and  d'd"c"c'd'  are  funicular  polygons 
for  them.  The  action  line  of  the  resultant  of  those  forces  passes 
through  D'  and  d\  the  intersections  respectively  of  the  extreme 
strings  of  those  polygons;  it  also  passes  through  Q  because  that 
point  is  the  limiting  position  of  the  extreme  strings  of  d'd"c"c'd' 
as  c"  or  d"  is  taken  nearer  and  nearer  to  P.  Hence  the  three 
points  are  on  a  straight  line,  which  fact  leads  to  the  following 
construction  for  the  projections  of  the  force  polygon: 

First  draw  the  projections  A'B'  and  A"B"  (of  F)  and  the 
indefinite  projections  B'C  and  B"C"  (of  FJ  and  A'D'  and  A"D" 
of  (F3) ;  then  guess  at  the  position  of  C,  as  at  c',  c" ,  and  supposing 
the  guess  to  be  correct,  draw  the  projections  c"d"  and  c'd'  of  F^\ 
next  prolong  A"d"  and  B"c"  to  their  intersection  P,  through 
which  draw  a  vertical  and  note  its  intersection  Q  with  B'C ;  finally 
draw  Q'd' ,  and  note  its  intersection  with  A'D'  (indefinite  in  length) — 
this  is  the  required  point  D'  from  which  the  projection  D'C  may 
be  drawn  and  then  D"C", 


INDEX. 

Kmnben  refer  to  pages. 


Acceleration,  17a,  186 

angular,  205,  208 
resolution  of,  188-X92 
Acceleration,  time  curves,  175 
Action  and  reaction,  3,  221 
Activity,  318 
Amplitude,  177 
Analysis,  methods  of,  2 
Angle  of  repose,  150 
Angular  acceleration,  205,  208 
displacement,  204,  208 
impulse,  331 
momentum,  333,  334 
velocity,  204,  208 
Attraction,  electric,  81-84 

gravitational,  74<-8t 
magnetic,  81-84 

Balancing,  279-283 
Blow,  force  of  a,  343,  344 

Catenary,  109 
Centrifugal  force,  261 
Centripetal  force,  261 
Centre  of  gravity,  defined,  54 

determination  of,  55-59 
Centre  of  mass,  220,  221 

of  percussion,  347,  348 

of  stress,  88 
Centrode,  213 
Centroid,  defined,  52,  60 

determination  of,  53,  60-73 
Collision,  344-347 
Components  of  a  force,  10-12,  14,  15 


j    Composition  of  couples,  49 
of  forces,  22-46 
of  forces,  defined,  10 
of  harmonic  motions,  196- 
202 
Compression,  86 
Conservation  of  energy,  316 
Cords,  flexible,  103- 113 
Couple,  defined,  9,  19 

graphical  representation  d,  20 
moment  of,  19 
resolution  of,  50 
Couples,  theory  of^  47-51 

D'AIeml^rt's  principle,  229 
Density,  75 

Dimensions  of  units,  357-362 
Displacement,  167,  177,  183 

angular,  204,  208 

defective,  298 
Dynamics,  defined,  i 
Dyne,  223,  226 

Effective  force,  228 
Efficiency,  317 

of  tackle,  324-326 
of  a  mine-hoist,  326,  32^ 
Energy,  306-313 

conservation  of,  316 
kinetic,  306-308 
mechanical,  312 
potential,  309-312 
principle  of,  for  machines,  317 
Equilibrium,  conditions  of,  93,  95-101 
defined,  93 

405 


INDEX, 


Field,  strength  of,  76,  82 
Force  at  a  distance,  3 
centrifugal,  261 
centripetal,  261 
components  of,  IO-12,  14,  15 
concentrated,  4 
contact,  3 
defined,  3 
distributed,  4 
effective,  228 
external,  93 

graphical  representation  of,  5 
internal,  93 
line  of  action  of,  4 
moment  of,  16,  17 
of  a  blow,  343 
parallelogram,  II 
polygon,  23 
.  resolution  of,  defined,  lO 
systems,  9 

transmissibility  of,  lO 
triangle,  il 
units  of,  6,  223 
Forces,  composition  of,  22-46 

composition  of,  defined,  lO 
concvurent,  defined,  9 
conservative,  311 
coplanar,  defined,  9 
non-concurrent,  defined,  9 
non-conservative,  311 
non-coplanar,  defined,  9 
resultant  of,  defined,  10 
Frequency,  177 
Friction,  149-158,  247,  248 
angle  of,  150 
belt,  157 
brake,  323 
circle,  154 

coefficient  of,  150,  151,  247 
cone,  156 
journal,  285,  286 
kinetic,  247,  248 
laws  of,  151,  247,  285 
pivot,  283 
rolling,  295-297 


Friction,  static,  defined,  149 
Funicular  polygon,  26,  398,  401 

Geekilogram,  224-226 
Geepound,  224-226 
Graphical  analysis,  2 
Gravitation  constant,  74 
law  of,  74 

Harmonic  motion,  composition  of,  196- 
202 

resolution  of,  198,  201 
simple,  177-180     ;  , 
Hinge  reactions,  275  ,   j      . 

Hodograph,  185 
Hoop  tension,  274 
Horse-power,  318 


Impact,  344-347 
Impulse,  329,  331 

angular,  33 1 
moment  of,  331 
sudden,  343 
Inertia  circle,  374-37^ 

ellipse,  374 
Instantaneous  axis,  212 
centre^  213 
rotation,  213 


Jet,  pressure  due  to  a,  340,  34I 

Kilogram,  6,  7 
Kinematics,  167 

defined,  i 
,  Kinetic  energy,  306-308 
friction,  247,  248 
reactions,  239,  271,  292 
units,  223 
Kinetics,  217 

defined,  i 

Laws  of  motion,  221,  223 
Lag,  177 
Lead,  177 


Numbers  refer  to  pages. 


INDEX. 


407 


Mass,  defined,  6,  219 

moment  of,  220,  249-257 
units  of,  6,  223-226 
centre,  220,  221 
Mechanics,   lefined,  i,  2 

subdivisions  of,  I 
Moment  of  a  couple,  19 

of  a  force,  16,  17,  395 
of  a  length,  61 
of  a  mass,  220,  249-257 
of  a  momentum,  333,  334 
of  a  volume,  61 
of  a  weight,  55 
of  an  area,  61,  363-378 
of  an  impulse,  331 
of  inertia,  249-257,  363-378 
of  inertia,   experimental  deter- 
mination of,  255 
of  inertia,  principal,  377,  378 
Moments,  principal  of,  28,  32,  39,  44. 
Momentum,  332-336 

angular,  333,  334 
moment  of,  333,  334 
Motion,  curvilinear,  183-202 
laws  of,  221,  223 
non-uniform,  168,  172,  184 
of  a  particle,  231-228 
of  a  rigid  body,  203-2 16,  233- 

297 
of  a  system  of  particles,    228- 

232 
plane,  207-216,  287-297 
rectilinear,  167-182 
relativity  of,  192-196 
uniform,  168,  176,  184 
uniformly  accelerated,  176 

Neutral  axis,  89 

Pappus  and  Guldinius,  theorem  o^  72 
Parallelogram  of  forces,  1 1 
Parallelepiped  of  forces,  13 
Pendulum,  ballistic,  347 
conical,  269 
Pendulums,  265-268,  269,  270 


Percussion,  centre  of,  347 

Period,  177 

Phase,  177 

Potential  energy,  309-312 

Pound,  6,  7 

Poundal,  224-226 

Power,  318 

Principal  axes,  257,  377 

Product  of  inertia,  257,  371-373 

Pulley,  113,  114,  325 

Radius  of  gyration,  250,  364 
principal,  377 

Rates,  351-356 

Repose,  angle  of,  150 

Resolution  of  acceleration,  188-192 
of  a  couple,  50 
of  a  force,  defined,  10 
of  harmonic  motion,  198,  20I 

Resultant  of  forces,  defined,  10 

Rigid  body,  defined,  228 

Rolling  resistance,  295-297 

Rotation,  203-207,  249-286 

Scalar,  349 

Shear,  86 

Space  diagram,  5 

Space-time  curve,  167,  183 

Speed-time  curve,  185 

Statics,  3 

defined,  I 
Stress,  85-92 

centre  of,  88 

diagram,  145 

intensity  of,  86 

Tackle,  115,  116 

efficiency  of,  324-326  - 
Tension,  86 
Torsion  balance,  269 
Train-resistance,  321 
Translation,  203,  233-248 
Transmissibility  of  force,  lO 
Triangle  of  forces,  11 
Trusses,  analysis  of,  136-148 


Numbers  refer  to  pages. 


4o8 


INDEX. 


Units,  absolute,  7,  223 
derived,  357 
dimensions  of,  357-362 
fundamental,  357 
gravitational,  6,  224 
kinetic,  223 

Varignon's  theorem,  17 

Vectors,  349,  350 

Vector  diagram,  5 

Velocity,  168,  184 

angular,  204,  208 
resolution  of,  188- 1 92 

Velocity-time  curve,  171 


Vibrations,  242-247 
Virtual  work,  381-386 

Watt,  318 

Weight,  apparent,  274 
defined,  8 
of  roofing,  387 
of  snow,  388 
of  trusses,  387 

Wind  pressure,  388 

Work,  298-305 

and  energy,  314,  32S 
diagram,  299 


Numbers  refer  to  pages. 


EXPLANATION   OF   PRINCIPAL  SYMBOLS. 


(The  numbers  refer  to  pages,  where  additional  explanation  may  be  found.) 

/  .  .  .  .length, 

M . .  .  moment  of  a  force 


A  .  .  .area. 

a .  . .  .  arm  of  a  force ;  linear  accel- 
eration. 

ax.  .  .arm  with  respect  to  x  axis; 
X  component  of  an  accelera- 
tion. 

a.  .  .  .acceleration  of  mass-centre. 

On.  .  .normal  component  of  accel- 
eration. 

at.  .  .tangential  component  of  ac- 
celeration. 

a.  ...angular  acceleration  (205); 
angle  of  repose  (150). 

C.  ..  .couple. 

Cx.  . .  component  of  C  perpendicular 
to  an  X  axis  (46). 

d.  .  .  .density. 
E.. .  . energy. 

Ek.  . .  kinetic  energy. 
Ep.  . .  potential  energy. 

e.  .  .  .efficiency  (317). 

e.  ...lead,    lag,    epoch,    or  epoch 

angle  (177,  197). 
F. . .  .  force ;    friction. 
F«.  ..X  component  of  a  force  (15). 
F' ...  limiting  friction  (150). 
<f>.  ..  .angle  of  friction  (150). 
/.. .  .moment  of  inertia  (250,  263). 
Ix-  .  .1  with  respect  to  x  axis. 
/.  . .  ./  with  respect  to  a  centroidal 

axis  (253,  366). 
/.  .  ..product  of  inertia  (257,  371). 
Jxy.  .  J  with  respect  to  x  and  y  axes. 
/.  .  . .  /  with  respect  to  central  axes 

(372). 
k.  . .  .radius  of  gyration  (250,  364); 
gravitation  constant  (74). 
kx.  .  .k  with  respect  to  x  axis. 
k.  .  .  .k  with  respect  to  a  centroidal 

axis. 


sometimes 
mass. 
Mq  .  .M  with  respect  to  O. 
m  .  .  .mass. 

N.  .  .normal  pressure  (149). 
n.  .  ..frequency  (177). 
P.. .  .force;    power  (318). 
p.  .  ..intensity  of  stress  (86). 
Q.  ..  .force. 

R. .  .  .resultant;    reaction. 
Rn.  ..normal  component  of  i?  (42, 

275)- 
i?<. .  .tangential   component    of  R 

(42,  275). 
r.  .  ..radius  to  mass-centre  (275). 
p.  . .  .radius  of  curvature;   surface 

density  (83). 

5.  .  . .  force. 

s.  .  .  .length  of  arc. 

r. ..  .tension;    period  (177). 

t.  .  .  .time. 

6.  ...angular  displacement  (204); 

angle. 
U.  .  .moment  of  momentum  (337). 
V  . .  .volume. 
V.  . . .  linear  velocity. 
Vx.  .  .X  component  of  v. 
V.  . .  .velocity  of  mass-centre. 
W . .  .  weight. 

IV.  ..  .work;    sometimes  weight. 
(o.  ..  .angular  velocity  (204);  solid 

angle  (79). 
X coordinate  of  centroid. 

y "        "       " 

z.  ...  ' 

Xc  . . .  coordinate  of  centre  of  stress 

(88). 
yc  . .  .coordinate  of  centre  of  stress 

(88). 

409 


V 


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